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UNIVERSITY   OF  CALIFORNIA. 

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MECHANICS  OF  THE  GIRDER:  • 


Treatise  on  ^Bridges  and  T^oofs, 


IN    WHICH   THE    NECESSARY  AND    SUFFICIENT    WEIGHT 

OF  THE   STRUCTURE   IS   CALCULATED, 

NOT   ASSUMED; 


THE  NUMBER  OF   PANELS  AND   HEIGHT  OF  GIRDER  THAT 

RENDER  THE   BRIDGE   WEIGHT  LEAST,  FOR  A 

GIVEN  SPAN,  LIVE  LOAD,  AND  WIND 

PRESSURE,  ARE  DETERMINED. 


BY 


JOHN   DAVENPORT  CREHORE,  C.E. 


'Inveniam  viam  out  faciam." 


NEW  YORK : 

JOHN     WILEY     AND     SONS. 
1886. 


COPYRIGHT,  1886, 
BY  MRS.  J.  D.  CREHORE. 

JJJJ 


RAND,  AVERY,  AND  CO., 

ELECTROTYPERS  AND  PRINTERS, 

BOSTON. 


PREFACE. 


THE  "Mechanics  of  the  Girder"  is  presented  to  the 
public  in  an  unfinished  condition,  just  as  it  was  left  at 
the  author's '  death,  in  October,  1884.  All  that  then  re- 
mained to  be  done  was  to  carry  out  an  example  in  each 
of  the  twelve  classes  of  girders  in  a  manner  similar  to 
that  of  the  Brunei  Girder  in  Class  I.  (Sections  2  and  3, 
Chapter  X.),  and  the  Double  Parabolic  Bow  and  Post  Truss 
in  Class  II.  (Chapter  XL).  Of  all  these,  the  Post  Truss 
promised  to  yield  the  most  prolific  results ;  and  it  may  be 
possible,  before  another  edition  is  published,  to  complete 
this  calculation  at  least,  if  not  to  introduce  other  examples 
from  the  later  classes.  However,  the  a  priori  method  of 
the  author  is  fully  set  forth  previous  to  the  tenth  chapter ; 
and  it  is  believed  that  no  one  else  has  as  yet  published 
any  so  satisfactory  results  from  this  method,  if,  indeed,  the 
method  has  been  hitherto  attempted  with  any  degree  of 
success. 

The  author's  family  feel  deeply  grateful  to  Professor 
John  N.  Stockwell  for  his  kindness  in  devoting  much  of 
his  valuable  time  to  the  supervision  of  the  proof-reading, 
for  the  many  suggestions  he  has  given  during  the  publi- 


iv  PREFACE. 


cation,  and  particularly  for  his  offer  to  conduct  the  work 
of  completing  the  remaining  examples.  At  his  own  sug- 
gestion, however,  it  has  been  thought  expedient  to  delay 
no  longer  the  publication  of  the  completed  portion  of  the 
book,  and  to  leave  any  additional  matter  to  be  inserted 
later. 

WILLIAM    W.   CREHORE. 

July  29,  1886. 


ANALYTICAL  TABLE  OF  CONTENTS. 


CHAPTER   I. 

PRESSURES   IN   ONE  PLANE. 
ART.  PAGE 

1.  Force  or  pressure  defined I 

2.  Resultant,  component,  equilibrium,  defined 2 

3.  The  parallelogram  of  forces 2 

4.  The  triangle  of  forces 3 

5.  The  resolution  of  a  force 4 

6.  The  resolution  of  many  forces  acting  in  one  plane  at  a  given  point    ...  4 

7.  The  composition  of  forces 5 

8.  Polygon  of  forces 6 


CHAPTER    II. 

MOMENT  OF  A   FORCE. 

9.  Definition  and  measurement  of  moment 9 

10.  Couples 9 

11.  Resultant  of  many  co-axal  couples 12 

12.  Arm  of  the  resultant  couple 14 

13.  Direction  of  the  resultant  couple 14 

CHAPTER   III. 

MOMENTS  OF  THE  EXTERNAL  FORCES  APPLIED  TO  A   BEAM  OR  GIRDER. 

SECTION  i. 
The  Semi-Beam,  or  Girder  fixed  at  One  End  and  free  at  the  Other. 

14.  Formulae  for  concentrated  and  for  uniformly  distributed  loads 17 

15.  Moments  due  many  equal  weights  placed  at  equal  intervals  along  the 

beam.     Examples 19 

1 6.  Moments  due  many  unequal  weights  placed  at  irregular  intervals      ...     22 

v 


VI  ANALYTICAL    TABLE   OF  CONTENTS. 


SECTION  2. 
Horizontal  Girder  supported  only  at  its  Ends,  which  are  not  fixed. 

ART.  PAGE 

17.  Data  for  required  formulae 23 

18.  Re-actions  at  the  piers 24 

19.  Tabulated  moments  for  any  load  or  pressure  on  the  girder 25 

20.  Moments   due   uniform   discontinuous   load   on   any   part   of   the   beam. 

Examples 27 

21.  Moment   at   the  foremost  end   of  a  uniform  load  advancing  by  panel 

weights.     Example 33 

22.  Moment  at  the  rth  weight  due  n  —  I  equal  weights  uniformly  distributed  .     34 

23.  Moment  at  foremost  end  of  advancing  uniform  load  when  the  two  end 

intervals  are  different  from  the  common  interval.     Example   ....     34 

24.  Moment  at  any  weight  due  n  equal  weights  applied  at  equal  intervals,  end 

intervals  different  from  common  interval.     Example 35 

25.  Difference  of  simultaneous  moments  at  consecutive  points  of  division. 

First  and  second  differences  of  these  simultaneous  moments.     Exam- 
ples.    Simple  computation  for  maxima  moments  and  differences     .     .     36 

26.  Point  of  greatest  moment  due  uniform  discontinuous  load.     Verification    .     40 

27.  Greatest  moment  at  foremost  end  of  advancing  uniform  load.     Example. 

Tabulated  results 41 

28.  General  expression  for  point  of  greatest  moment  due  uniform  dead  and 

advancing  uniform  live  load 44 

29.  Moment  at  foremost  end  of  continuously  distributed  uniform  load,  advan- 

cing by  continuous  increments,  and  not  by  leaps.     Example    ....     45 

30.  Moment  due  both  dead  and  live  loads  one  interval  beyond  the  foremost 

end  of  advancing  load.     Convenient  method  of  computation    ....     46 

31.  Position  of  foremost  end  of  uniform  continuous  live  load  when  moment  at 

that  end  is  a  maximum 48 

32.  Position  of  foremost  end  of  same  when  moment  there  is  maximum  for 

combined  dead  and  live  loads 48 

33.  Maximum  moment  due  any  uniform  partial  or  complete  continuous  load   .  48 

34.  Point  of  greatest  moment  due  full  continuous  uniform  load 49 

35.  Point  of  greatest  moment  due  both  dead  and  live  loads 49 

36.  Moments  when  the  panels  are  equal,  but  the  weights  are  unequal,  and 

are  not  applied  at  equal  intervals.     Example.     Two  locomotives     .     .     49 

37.  Moments  due  the  same  locomotives  when  their  weight  is  uniformly  dis- 

tributed throughout  their  length 57 

SECTION  3. 
Horizontal  Girder  of  One  Span,  with  Fixed  Ends.    Effects  of  End  Moments. 

38.  Formula  for  computing  the  influence  of  the  end  moments  on  the  normal 

moments 59 

39.  Formula  for  the  correction  of  the  normal  differences  of  moment  for  influ- 

ence of  end  moments.     Examples  illustrative 60 


ANALYTICAL    TABLE   OF  CONTENTS,  vii 


CHAPTER    IV. 

STRAINS    IN    FRAMED    OR   BUILT   GIRDERS,  DEDUCED   FROM   THE   MOMENTS 
OF   THE    EXTERNAL   FORCES   AND   FROM   THE   SHEARING-FORCES,    AND 

FROM   THESE  COMBINED. 
ART.  PAGE 

40.  Statement  of  a  case 64 

41.  Strains  found 66 

42.  Shearing-forces  and  shearing-strains.     Definitions,  with  Table  1 67 

43.  Expressions  for  shearing  force  and  strain 69 

44.  Formulae  for  increments  of  shearing-strains  due  to  end  moments  ....  73 

45.  Shearing-strain  at  any  vertical  section  of  a  girder  in  terms  of  the  vertical 

components  of  the  forces  which  are  impressed  upon  the  shearing-plane 

through  the  members  of  the  girder  cut  by  that  plane 74 

46.  Strains  in  all  members  of  a  girder  determined  from  the  given  shearing- 

forces      .     .     .     .     i 75 

47.  Examples  illustrative  of  two  methods  of  determining  strains  in  open  girders  .  76 

48.  Maxima  strains  in  the  web  members  of  an  open  girder,  deduced  from  the 

moments  and  shearing-forces  combined,  for  uniform  discontinuous  dead 

and  live  loads 80 

49.  Classification  of  girders 86 


CHAPTER  V. 

MOMENTS    OF    RESISTANCE    OF    THE    INTERNAL    FORCES    OF    A    BEAM    OR 
GIRDER   HAVING  A  CONTINUOUS   WEB. 

SECTION  i. 

General  Formula  found  and  applied  to  Particular  Cross-Sections  of^  Beams 
with  Continuous  Web. 

50.  Mode  of  estimating  the  moment  of  resistance  . 134 

51.  Beam  of  rectangular  cross-section 136 

52.  Hollow  beam  of  rectangular  cross-section 137 

53.  Beam  composed  of  two  vertical  plates  and  two  horizontal  channels  .     .     .138 

54.  Beam  composed  of  two  vertical  I-beams  and  two  equal  horizontal  plates  .  139 

55.  Beam  composed  of  two  vertical  channels  and  two  horizontal  plates  .     .     .  140 

56.  Beam  of  T-shaped  section 140 

57.  Solid  or  hollow  square  beam  with  diagonal  vertical  compared  with  the 

same  beam  when  a  side  is  vertical     .'..... 142 

58.  Solid  or  hollow  beam  of  circular  cross-section <  144 

59.  Solid  or  hollow  beam  of  elliptical  cross-section 145 

60.  Relation  among  the  ultimate  resistances  of  material  to  tension,  to  com- 

pression, and  to  cross-breaking ;  with  Table  II 146 


viii  ANALYTICAL    TABLE   OF  CONTENTS. 


SECTION  2. 
Moment  of  Inertia  and  Radius  of  Gyration  of  a  Given  Cross-Section. 

ART.  PAGE 

61.  Definitions 154 

62.  Eprmulae  for  moment  of  inertia  and  the  square  of  radius  of  gyration. 

Table  III 154 


CHAPTER  VI. 

DEFLECTION,    END    MOMENTS,    AND    POINTS    OF   CONTRARY    FLEXURE 
FOUND.  —  CAMBER. 

SECTION  i. 
Deflection  of  the  Semi- Beam  having  a  Uniform  Cross-Section. 

63.  Equation  of  the  elastic  curve  as  applied  to  a  beam  or  pillar 157 

64.  Deflection  of  the  semi-beam  under  one  weight 158 

65.  Deflection  of  the  semi-beam  under  uniform  load  per  unit 159 

66.  Semi-girder,  partial  uniform  load  on  each  unit  of  length  at  any  distance 

from  fixed  end 160 

67.  Total  deflection  of  semi-beam  at  its  free  extremity  when  supporting  a  load 

not  reaching  that  extremity.     Example 162 

68.  Total  deflection  of  semi-beam  at  its  free  extremity  when  supporting  equal 

weights  at  equal  intervals,  and  last  weight  is  at  the  free  end    .     .     .     .164 

69.  Total  deflection  of  semi-beam  at  its  free  extremity  when  weight  at  free  end 

is  one-half  every  other  weight 166 

70.  Deflection  at  a  given  interval  due  to  all  the  equally  distributed  equal 

weights 166 

71.  Deflection  at  a  given  interval  when  weight  at  the  free  end  is  wth  part  of 

every  other.     Example 167 

SECTION  2. 

Deflection  of  a  Beam  of  Uniform  Cross-Section,  supported  at  its  Free  Unflxed 

Ends. 

72.  Deflection  due  the  beam's  own  weight  supposed  to  be  uniform.    Example  .  169 

73.  Deflection  due  a  concentrated  load  placed  at  a  given  horizontal  distance 

from  the  end  of  the  beam.    Example 170 

74.  Deflection  due  a  uniformly  distributed  partial  load 174 

75.  Application  to  the  special  case  of  an  advancing  continuous  load.    Examples.  178 

76.  Deflection  at  any  point  due  any  number  of  equal  weights  placed  at  equal 

intervals  along  the  beam.    Examples 183 


ANALYTICAL    TABLE   OF  CONTENTS.  ix 


SECTION  3. 

The  Influence  of  Fixed  Ends  upon  the  Deflection  of  a  Beam  of  Uniform 
Cross-Section,  supported  at  its  Two  Extremities,  which  are  assumed  to 
be  Level,  and  One  or  Both  of  Them  fixed  Horizontally  or  Otherwise. 
Determination  of  the  End  Moments  and  Points  of  Contrary  Flexure. 

ART.  PAGE 

77.  Deflection  due  end  couples 189 

78.  Deflection  at  any  point   of   beam,  with  load  continuous   and  uniform 

throughout '. 190 

79.  Deflection  of  beam  horizontally  fixed  at  one  end,  and  simply  supported  at 

the  other.     Example 192 

80.  Deflection  of  beam  fixed  horizontally  at  both  ends,  due  to  a  concentrated 

load  placed  at  a  given  distance  from  the  left  end  of  beam 194 

81.  Points  of  maximum  deflection  and  points  of  contrary  flexure.     Examples  .  197 

82.  Deflection  due  any  number  of  equal  weights  placed  at  equal  intervals 

along  beam  fixed  horizontally  at  both  ends 201 

83-  Deflection  of  beam  fixed  at  right  end,  and  simply  supported  at  the  left, 

uniformly  loaded 207 

84.  Deflection,  end  moments,  and  points  of  contrary  flexure  due  a  partial  uni- 

form load  continuously  distributed  when  both  ends  of  the  beam  are 
fixed  horizontally 209 

85.  Partial  or  full  continuous  uniform  load  on  any  portion  of  a  beam  fixed 

horizontally  at  the  right  end,  but  simply  supported  at  the  left .     .     .     .  213 

86.  Examples  illustrative 214 

87.  Established  formulae  applied  when  intervals  are  fractional,  not  integral. 

Examples 219 

88.  Continuous  uniform  load  on  beam  fixed  horizontally  at  both  ends     .    .     .221 

SECTION  4. 

Deflection  of  a  Girder  of  Variable  Cross-Section  in  Terms  of  the  Constant 
Unit  Strain  upon  the  Extreme  Fibres  of  the  Section  ;  that  is,  Deflection 
of  a  Beam  of  Uniform  Strength.  End  Moments  for  Fixed  Beams. 

89.  How  cross-sections  of  members  should  be  proportioned.     Equation  used  .  224 

90.  Deflection  of  semi-girder  of  uniform  height  and  strength.     Example      .     .  224 

91.  Deflection  of  the  semi-girder  of  uniform  strength  but  of  variable  height. 

Examples   . 226 

92.  Additional  applications 235 

93.  Deflection  of  the  girder  of  uniform  strength,  supported  at  both  ends,  either 

fixed  or  free,  and  the  height  of  the  girder  being  either  uniform  or  vari- 
able.    Examples 235 

94.  Preceding  results  arranged  according  to  amount  of  deflection 240 

95.  Total  deflection  nearly  in  the  inverse  ratio  of  the  areas  of  the  figures  of 

the  girders i 241 


ANALYTICAL    TABLE   OF  CONTENTS. 


96.  These  formulae  may  be  employed  for  girders  of  continuous  web.  Examples  .  243 

97.  Thickness  of  continuous  webbed  girder.     Example 246 

98.  Thickness  of  the  beam  at  any  point.     Example 248 

99.  Determination  of  any  dimension  of  cross-section.     Example 249 

100.  Beam  of  uniform  strength  fixed  horizontally  at  both  ends 251 

101.  Beam  of  uniform  strength  fixed  horizontally  at  both  ends;  points  of  con- 

trary flexure 252 

102.  Beam  of  uniform  strength  and  height  fixed  at  both  ends,  and  bearing  a 

concentrated  weight.     Examples 252 

103.  Beam  of  uniform  strength,  height,  and  load,  fixed  horizontally  at  both 

ends.     Rectangular  cross-section.     Example 256 

104.  Concentrated  weight  at  given  distance  from  the  left  end  of  a  beam  of  uni- 

form strength,  fixed  at  the  right  end  only.     Example 258 

105.  Continuous  uniform  load  on  a  beam  of  uniform  strength,  fixed  at  the  right 

end  only.     Example ,  261 

106.  Beam  of  uniform  strength  and  uniformly  varying  height,  fixed  at  both  ends. 

Examples 263 

.107.  Beam  of  uniform  strength  and  uniformly  varying  height,  fixed  at  one  end. 

Examples 273 

SECTION  5. 
Camber. 

108.  Definition.     Modes  of  giving  camber 277 

109.  Change  of  length  calculated  from  the  unit  strain 278 

no.  Elongation  and  contraction  calculated  from  deflection  .          279 

in.  Application  to   Classes  II.,  IV.,  VII.,  IX.,  X.,  and   XII.  of  article  49. 

Example 281 

112.  Applied  to  the  straight  upper  horizontal  chord  of  Classes  III.,  IV.,  VIII., 

IX.,  XI.,  XII 284 

Ji3.  Treatment  for  change  of  length  in  chords  and  verticals.     Length  of  diag- 
onals changed.     Example 286 


CHAPTER  VII. 

THE  STRENGTH   OF  PILLARS,   COLUMNS,    POSTS,   OR   STRUTS. 

SECTldN    I. 
Strength  of  Pillars  by  Rational  Formula. 

114.  Definitions 289 

115.  Pillars  of  uniform  cross-section ;  three  cases 290 


ANALYTICAL    TABLE   OF  CONTENTS.  xi 

SECTION  2. 

Hodgkinsorfs  Empirical  Formula:  for  the  Strength  of  Cast-Iron  and  Timber 

Pillars. 

ART.  PAGE 

116.  Formulae  for  cast-iron  pillars 296 

117.  Formulae  for  timber  pillars 298 

SECTION  3. 
Gordon's  Empirical  Formula,  with  Rankine^s  Modification. 

118.  Gordon's  formula  deduced 298 

119.  Rankine's  modification  of  same,  introducing  the  least  radius  of  gyration 

of  the  cross-section 300 

1 20.  Explanation  of  variations.    Table  IV 300 

SECTION  4. 

Strength  of  Pillars  compiited  by  the  Preceding  Formula^  and  compared  with 
the  Strength  experimentally  determined. 

121.  Tables  V.,  VI.,  VII.,  VIIL,  IX.,  X.,  XL,  XII 303 


CHAPTER  VIII. 

PROPORTIONS  AND  WEIGHTS   OF  ALL  THE   MEMBERS  OF  A   BRIDGE 
EXCEPTING  THE  GIRDERS  PROPER. 

122.  The  floor 312 

123.  The  joists,  longitudinal 312 

124.  The  wrought-iron  I  floor  beams,  transverse 314 

125.  The  system  of  lateral  support 316 

126.  Additional  weight 317 

127.  Example 317 


CHAPTER    IX. 

OPEN  GIRDERS  WITH  EQUAL  AND  PARALLEL  STRAIGHT  CHORDS.     CLASS  IX. 

SECTION  i. 
The  Pratt  Truss  of  Single  System  and  Uniform  Live  Load.  —  Wind  Pressure. 

128.  Strains  in  terms  of  the  structure's  unknown  weight 318 

129.  Weight  of  the  structure  determined 324 

130.  Bridge  of  two  equal  girders  with  additional  permanent  weight.      Strain 

sheet 330 


Xll  ANALYTICAL    TABLE   OF  CONTENTS. 


ART.  PAGE 

131.  Discussion  of  preceding  strain  sheet 333 

132.  Changes  to  be  made  in  strain  sheet 334 

133.  Lateral  sway-bracing 337 

134.  Best  number  of  panels,  and  best  height.     Preliminary  operations .     .     .     .337 

135.  Strains  due  to  wind.     Weight  of  beams 338 

136.  Strains  due  to  wind.     Diagonal  strains 341 

137.  Lateral  strengthening  of  end  posts 343 

138.  Additional  strains  in  end  members  required  to  resist  wind  pressure  .     .     .  343 

139.  Formulae  for  weight  to  be  added  to  resist  adjustment  and  distortion  strains  .  348 

140.  Exemplification  of  method  of  article  138 351 

141.  Exemplification  of  method  of  article  139 365 

142.  Exemplification  of  method  of  article  139  with  double  the  load 373 

143.  Another  example 382 

144.  Deductions  and  comparisons 391 

SECTION  2. 

The  Pratt  Truss  of  Single  System  under  Varying  Live  Load,  without  taking 
Account  of  Wind  Pressure. 

145.  Example  of  the  two  locomotives  again 392 

146.  For  deck  bridge 398 

147.  Col.  Merrill's  example  for  Pratt  Truss 399 


•CHAPTER  X. 

CALCULATION  OF  THE  WEIGHT  OF  BRIDGES  HAVING  GIRDERS  OF  CLASS  I., 
AND  DETERMINATION  OF  THE  NUMBER  OF  PANELS  AND  THE  HEIGHT 
OF  GIRDER,  WHICH  RENDER  THE  BRIDGE  WEIGHT  LEAST  FOR  A  GIVEN 
SPAN  AND  UNIFORM  LIVE  LOAD.  —  LIMITING  SPAN  FOUND. 

SECTION  i. 

General  Specifications  for  Iron  Bridges,  issued  in  1879  ty  the  New  York, 
Lake  Erie,  and  Western  Railroad  Company.  O.  Chanute,  Chief 
Engineer. 

148.  General  specifications  for  iron  bridges ,....413 

SECTION  2. 
The  Brunei  Girder  of  Single  System. 

149.  Dimensions  computed 423 

150.  Moments  and  strains  in  chords  due  to  the  total  panel  weight 427 


ANALYTICAL    TABLE   OF  CONTENTS.  Xlll 


ART.  PAGE 

151.  Weights  of  these  wrought-iron  chords 429 

152.  Greatest  strains  and  weights  in  the  web  system 433 

153.  Weight  of  floor 440 

154.  Weight  of  joists 441 

155.  I  floor  beams 441 

156.  More  convenient  method  for  weight  of  joists 442 

157.  Weight  of  wrought-iron  I-beams 443 

158.  Transverse  I-beams 444 

159.  Horizontal  wind  pressure '.     .     .     .     .  446 

160.  The  horizontal  diagonals 446 

161.  The  horizontal  struts;   that  is,  in  this  case,  the  quantity  of  iron  to  be 

added  to  the  transverse  I-beams  by  reason  of  wind  pressure    ....  448 

162.  The  chords  required  in  the  horizontal  system  to  resist  wind  force     .     .     .  450 

163.  Vertical  supports 451 

164.  Lateral  head  bracing  as  additional  security  against  deflection 455 

165.  Necessary  amount  of  material  for  the  triangular  web  system  of  latticed 

struts  or  columns 459 

166.  Weight  of  the  bridge 462 

167.  Table  showing  the  number  of  panels  and  height  which  simultaneously  ren- 

der the  total  bridge  weight  a  minimum 466 

168.  Inferences  from  table 471 

169.  Example.     Strain  sheet 472 


SECTION  3. 
The  Brunei  Girder  of  Double  System. 

170.  Expression  for  height 485 

171.  Weight  of  chords 486 

172.  Weight  of  girder  diagonals 488 

173.  Whole  weight  of  girders  due  to  loads 490 

174.  Weight  of  floor 490 

175.  Weight  of  longitudinal  I-beams 491 

176.  Weight  of  transverse  I-beams 492 

177.  Weight  added  to  transverse  I-beams  for  wind 492 

178.  Weight  of  horizontal  diagonals 493 

179.  Weight  of  wind  chords      . 493 

180.  Weight  of  all  verticals 494 

181.  Weight  of  head  system 496 

182.  Equations  and  table  giving  best  height  for  least  bridge  weight.     Two 

Brunei  Girders  of  double  system 501 

183.  Inferences  from  table 512 

184.  Example.     Strain  sheet 512 


xiv  ANALYTICAL    TABLE   OF  CONTENTS. 


CHAPTER   XL 

BRIDGES  OF  CLASS  II.  —  BEST  NUMBER  OF  PANELS  AND  BEST  HEIGHT 
DETERMINED  FOR  A  GIVEN  SPAN  UNDER  A  GIVEN  UNIFORM  LIVE 
LOAD,  —  LEAST  BRIDGE  WEIGHT  AND  LIMITING  SPAN  FOUND. 

SECTION  i. 

The  Parabolic  Bowstring  Girder  of  Double  Triangular  System,  with  the 
Extreme  Diagonals  omitted^  and  a  Vertical  Suspender  at  Extreme 
Panel  Point. 

ART.  PAGE 

185.  Dimensions 523 

186.  Weight  of  chords 524 

187.  The  girder  diagonals 526 

188.  Floor 532 

189.  Weight  of  longitudinal  I-beams 532 

190.  Weight  of  transverse  I-beams 532 

191.  Weight  of  horizontal  diagonals 532 

192.  Weight  of  wind  chords 532 

193.  Weight  of  verticals 534 

194.  The  head  lateral  system 535 

195.  Equations  and  table  giving  best  height  and  best  number  of  panels  for  least 

bridge  weight 535 

196.  Inferences  from  table 540 

197.  Example.     Strain  sheet 540 

SECTION  2. 
The  Post  Truss  with  Parabolic  Top  Chord. 

198.  Dimensions  and  expressions  for  height 546 

199.  Moments  due  a  total  dead  load  of  uniform  panel  weight 549 

200.  Weights  of  top  and  bottom  chords  due  a  total  dead  load  of  uniform  panel 

weight    . 558 

201.  Greatest  strains  in  the  girder  diagonals,  and  their  weights 563 


MECHANICS  OF  THE  GIRDER, 


CHAPTER   I. 

PRESSURES    IN   ONE    PLANE. 

i.  FORCE  is  a  cause  which  changes,  or  tends  to  change,  the 
condition  of  matter  as  to  rest  or  motion.  Whether  there  is  or 
is  not,  in  fact,  any  difference  between  force  and  pressure,  it  is 
sufficient  for  the  purposes  of  this  volume  to  treat  them  as 
identical,  since  it  is  with  their  measurable  effects  alone  that 
we  are  here  concerned. 

A  force  is  said  to  be  given  when  its  point  of  application,  its 
direction,  its  line  of  action,  and  its  intensity  are  known.  Two 
pressures  are  equal  which,  acting  on  the  same  point,  along  the 
same  line,  and  in  opposite  directions,  neutralize  each  other ; 
and,  if  two  equal  pressures  act  at  the  same  point  in  the  same 
direction,  the  result  of  their  combined  action  is  twice  that  of 
each  separate  pressure. 

Pressures,  therefore,  may  be  compared  by  means  of  numbers 
expressing  their  intensities.  Since  the  intensity  of  any  one  of 
the  pressures  to  be  compared  may  be  taken  as  the  standard,  it 
follows  that  the  unit  pressure  is  entirely  arbitrary,  and  may  be 
a  finite  or  an  infinitesimal  pressure. 

When  pressures  are  expressed  by  symbols,  such  as  P,  Q,  R> 
etc.,  it  is  to  be  understood  that  these  letters  stand  for  num- 


MECHANICS  OF   THE   GIRDER. 


bers  denoting  the  number  of  times  the  concrete  unit  is  taken. 
Otherwise,  such  an  expression  as  P2,  being  the  square  of  a 
concrete  pressure,  would  be  unintelligible. 

A  force  or  pressure  may  be  conveniently  represented  by 
a  geometrical  straight  line ;  one  end  of  the  line  denoting  the 
point  of  application  of  the  force,  the  direction  of  the  line  being 
coincident  with  the  direction  of  the  force,  and  the  number  of 
linear  units  in  the  line  being  equal  to  the  number  of  force  units 
to  be  represented. 

2.  When  many  pressures  act  at  the  same  time  on  a  mate- 
rial particle,  the  result  of  their  combined  action  is  generally 
a  definite  pressure  in  a  definite  direction.     This  definite  press- 
ure is  called  the  resultant  of  the  acting  or  impressed  pressures ; 
and  these  latter,  with  reference  to  the  resultant,  are  styled  com- 
ponents.    When  the  resultant  is  zero,  the  pressures  are  said  to 
be  in  equilibrium ;  when  the  resultant  of  the  given  pressures 
is  not  zero,  equilibrium  may  evidently  be  produced  by  intro- 
ducing a  new  force  which  shall  neutralize  this  resultant. 

3.  Parallelogram  of  Forces It  is  shown  in  elementary 

works  on  mechanics,  that  if  two  forces  act  upon  a  single  point, 

and  their  intensities  and  direc- 

A ^> 

tions  be  represented  by  two  ad- 
jacent sides  of  a  parallelogram, 
then  the  diagonal  of  the  paral- 
lelogram drawn  to  the  intersec- 
tion of  those  two  sides  will 

represent,  both  in  magnitude  and  direction,  the  resultant  of  the 
two  given  forces. 

If  Pt  and  P2,  Fig.  i,  are  two  forces  acting  at  the  point  <9, 
represented  in  magnitude  and  direction  by  the  lines  OA  and 
OB,  then,  completing  the  parallelogram,  AOBC,  the  resultant 
will  be  represented,  in  magnitude  and  direction,  by  the  diagonal 
OC  —  R.  When,  therefore,  a  force  is  applied  at  O  equal  in 
intensity  to  R,  and  acting  in  the  same  line  but  in  the  opposite 


PRESSURES  IN  ONE  PLANE. 


direction,  it  will  balance  the  given  forces  Plt  P2,  and  the  three 
forces  will  be  in  equilibrium. 

4.  Triangle  of  Forces.  —  Since,  in  Fig.  i,  AO  =  BC  =  Piy 
the  three  sides  of  the  triangle  BOC  (or  A  OC)  represent,  in  mag- 
nitude and  direction,  three  forces,  Pu  P2,  and  R,  which,  acting 
in  one  plane  on  a  given  point,  are  in  equilibrium  ;  the  direction 
of  the  forces  being  that  of  a  point  traversing  the  perimeter  of 
the  triangle.  In  this  manner  the  value  of  the  resultant  may  be 
constructed. 

A  formula  for  the  value  of  R  is  found  by  solving  the  tri- 
angle of  forces,  where  two  sides  and  the  angle  included 
between  them  are  given.  Thus,  if  c  =  AOB  =  the  angle 
between  the  given  lines  of  action  of  Pl  and  P2,  we  have,  from 
the  geometry  of  the  figure,  putting  BOC  —  6  (tketa), 

R2  =  P*  +  P22  +  2P1P2  cos;-.  (i) 


Sin  0  =        sin  <:.  (2) 


EXAMPLE.  —  Let  P,  =  8,  P2  =  12,  c  =  75°. 
Then 

R*  =  82  +  i22  +  2  x  8  x  12  cos  75°  =  208  +  192  X  0.25882 

=  2S  7. 
R   =  16.053. 


Sin#  =  -  x  0.96593  =  0.48471. 
16.053 

e  =  28°  59'  4o". 

If  the  lines  of  action  of  the  two  forces,  P19  P2,  are  at  right 
angles  to  each  other,  cos^  becomes  zero,  and  equation  (i)  re- 
duces to  R2  =  P2  4-  P22,  where  R  is  the  hypothenuse,  and  Pl 
and  P2  are  the  other  sides  of  a  right-angled  triangle. 


MECHANICS  OF  THE   GIRDER. 


EXAMPLE.  —When  Px  =  8,  and  P2  =  12, 

R2  =  82  +  i22  =  208.        R  =  14.422. 
In  this  case,  Fig.  2,  we  have 


>  =  R  sin  0  =  P2  tan  (9. 

2  —  R  cos  0  =  />,  cot  (9. 

=  jP,  -*-  sin  0  =  Pt  cosec  0. 

?  =  P2  -5-  cos  0  =  P2  sec  ^. 


(3) 


FIG.  2. 

5.  Resolution  of  a  Force.  —  Conversely,   any  force,   R, 
acting  at  a  given  point  with  known  intensity  and  direction, 
may  be  resolved  into  two  component  forces  acting  at  the  same 
point,   having  definite  intensities   and  directions.     Manifestly 
also  may  each  one  of  the  two  components  be  resolved  into 
two  components,  and  so  on  without  limit. 

EXAMPLE.  —  Resolve  the  force  R  =  100  tons,  acting  at  the 
point  Oy  Fig.  2,  in  the  direction  OC,  into  its  horizontal  and 
vertical  components  ;  6  being  equal  to  28°  59'  40". 

From  (3), 

/!  =  R  sin  0  =  100  x  0.48471  =  48.471  tons. 

P2  =  R  cos0  =  100  X  0.87467  =  87.467  tons. 

6.  Resolution  of  Many  Forces  acting  in  One  Plane  at  a 
Given   Point.  —  Let  there  be  any  number  of  forces,  Pa  P2, 
P3,  etc.,  Fig.  3,  acting  in  the  plane  of  the  axes  XX',  YY',  at 
their  point  of  intersection,  O ;  and  let  a  (alpha)  symbolize  the 
angle  between  the  line  of  action  of  any  force  and  the  axis  of  x. 


PRESSURES  IN  ONE  PLANE. 


Resolving  each  force  into  its  horizontal  and  vertical  com- 
ponents, and  calling  the  sum  of  the  horizontal  components  X> 
and  the  sum  of  the  vertical  components  F,  these  results  : 


X  =  Pl  cos  <*r  4-  P2  cos  a2  +  P3  cos  «3  +  .  .  .  =  2/>cos  «,      (4) 
Y  —  PI  sin  «x  +  P2  sin  «2  •+-  />3  sin  a3  +  .  .  .  =  SPsin  a  ;     (5) 


the  symbol  2  (sigmd)  denoting  the  sum  of  the  terms  having  the 
form  P  cos  a  or  P  sin  a. 

Y 


7-  Thus,  for  all  the  given  forces  acting  in  their  various 
directions  on  the  point  O  have  been  substituted  two  other 
forces,  X  and  F,  acting  at  the  same  point ;  the  one  horizon- 
tally, the  other  vertically,  and  in  the  plane  of  the  original 
forces.  Now,  if  R  is  the  resultant  of  the  two  forces  X  and  F, 
it  must  also  be  the  resultant  of  the  forces  P»  P2,  Py  etc. ;  and, 
0  being  the  angle  between  the  resultant  and  the  axis  of  x,  we 
shall  have 

Rcos9     =  X  =  IPcosa,  (6) 

R  sin  6     =  F  =  XPsin  «,  (7) 

.-.    fr  =  X2  +  Y*,  (8) 

=  -V  (9) 


6  MECHANICS  OF   THE   GIRDER. 

When  the  given  forces  are  in  equilibrium,  the  resultant 
vanishes,  and 

X  =  ZPcosa  =  o.  (10) 

Y  =  2/>sin  a  =  o.  (u) 

EXAMPLE.  —  Let  PI  =  10  tons,  a,  =    40°. 

P2  =  20  tons,  a2  r=  150°. 

P3  =  30  tons,  a3  =  250°  —  —  1 10°. 

P4  =  40  tons,  a4  =  300°  =    —60°. 

Required  the   intensity,  R>  and   the  direction,   0,  of   the  re- 
sultant. 

X  =  10  cos  40°  +  20  cos  150°  +  30  cos  250°  4-  40  cos  300°  =  10  cos  40° 

—    20  COS  30°  —  30  COS  70°  4-  40  COS  60°  =   10   X   0.76604  —  20 

X  0.86603  —  30  x  0.34202  4-  40  x  0.5  =  0.0792  tons. 
Y  =  10  sin  40°  4-  20  sin  150°  4-  30  sin  250°  -f-  40  sin  300°  =  10  sin  40° 
+  20  sin  30°  —  30  sin  70°  —  40  sin  60°  =  10  x  0.64279  +  20 
X  0.5  —  30  x  0.93969  —  40  x  0.86603  =  —46.404  tons. 


0.0792 
=  [(o.o792)2  4-  (— 46.404)2]s  =  Y -t-  sin0  =  46.404065  tons. 


The  resultant  is  therefore  in  the  fourth  quadrant,  and  makes  an 
angle  of  5'  52"  with  the  axis  of  y. 

This  substitution  of  one  force  for  many  others  is  called  the 
composition  of  forces. 

8.  Polygon  of  Forces.  —  Let  St,  S2,  53,  etc.,  in  Fig.  4,  be 
the  five  sides  of  a  closed  polygon.  Measure  the  inclination  of 
each  side  to  the  horizon,  as  indicated  in  the  figure,  for  ciy  c2) 
etc. ;  then  the  sum  of  the  horizontal  projections  of  all  the 


PRESSURES  IN  ONE  PLANE. 


sides  is,  in  accordance  with  the  trigonometrical  signs  of  the 
cosine,  found  to  be 


Since 


cos  c  =  Si  cos  <T!  +  S2  cos  c2  -J-  S3  cos  cz  4-  S4  cos  c4 

4-  ,SS  cosrs  =  o.     (12) 


+BC,  S2  cos<r2  =  —  DE, 

cos<rs  =  -\-ABy  S3  cos^r3  =  —FG, 


FIG.  4. 

Equation  (12)  is  true  whatever  be  the  number  of  sides  of 
the  polygon;  and  from  its  analogy  to  equation  (10),  viz., 

2Pcosa  =  o, 

we  may  enunciate  the  proposition,  that  when  any  number  of 
forces  acting  at  the  same  point,  with  their  lines  of  action  in 
the  same  plane,  are  in  equilibrium,  then  the  given  forces  may 
be  represented,  in  magnitude  and  direction,  by  the  sides  of  a 
closed  polygon  ;  the  direction  being,  for  each  side,  that  of  a 
point  traversing  the  perimeter. 

This  proposition  enables  us  to  construct  the  resultant  of 
many  forces  acting  on  a  point  in  the  common  plane  of  their 


8 


MECHANICS  OF  THE   GIRDER. 


x' 


lines  of  action,  by  regarding  the  unknown  resultant,  with  its 
direction  changed,  as  the  side  required  to  complete  or  close  the 
polygonal  figure  due  to  the  given  forces. 

EXAMPLE.  —  Let  us  apply  this  proposition  to  the  example 
of  Art.  7. 

This  solution  consists 
simply  in  drawing  (Fig.  5) 
a  continuous  figure  made 
up  of  lines  proportional  to 
the  given  forces  and  re- 
spectively parallel  to  their 
given  lines  of  action,  and 
each  force  having  the  di- 
rection a  point  would  take 
in  traversing  the  broken 
line  from  end  to  end.  The 
straight  line  joining  the 
two  ends  of  this  broken 
line  will  be  the  resultant 
sought,  with  its  direction 
reversed. 

It  will  be  seen  that  the 
values  of  Y  and  R  in  this 
example  are  very  nearly 
equal,  and  that  the  solution 
by  construction  can  show 
an  appreciable  difference 


in  them,  only  when  a  large  scale  is  used.  In  practice,  however, 
either  solution  is  accurate  enough  ;  and  one  serves  to  check  the 
other. 

The  triangle  of  forces  is  a  particular  case  of  the  closed 
polygon. 


MOMENT  OF  A   FORCE.  9 


CHAPTER  II. 

MOMENT   OF    A    FORCE. 

9.  The  moment  of  a  force  is  the  effect  of  the  force's  effort 
to  turn  the  body  on  which  it  acts  about  a  given  point,  and  is 
measured  by  the  number  expressing  the  force,  multiplied  by 
the  number  denoting  the  perpendicular  distance  from  the  given 
point  to  the  line  of  action  of  the  force. 

Moments,  therefore,  may  be  added  and  subtracted,  and 
represented  by  lines,  like  other  numbers. 

Since  the  unit  of  the  force  and  the  unit  of  the  perpendicular 
distance  are  arbitrary,  it  is  usual  to  express  the  moment  as  a 
denominate  number,  designating  both  the  units.  Thus,  20 
foot-tons,  or  ton-feet,  means  that  the  moment  20  is  equivalent 
to  the  effect  of  a  force  or  pressure  of  20  tons  acting  at  the  per- 
t  pendicular  distance,  or  lever  arm,  of  i  foot  from  the  axis  of 
rotation,  or  to  a  force  of  i  ton  acting  at  the  distance  of  20  feet 
from  the  same  axis. 

It  is  plain  that  the  moment  of  a  given  force  acting  at  a 
given  perpendicular  distance  from  the  axis  of  rotation  may 
be  replaced  by  any  one  of  an  infinite  number  of  equivalent 
moments. 

10.  In  former  articles  forces  have  been  considered  as  acting 
in  straight  lines  in  one  plane  and  on  a  single  point ;  tending 
in  their  united  action,  when  the  resultant  does  not  vanish,  to 
move  that  point  or  material  particle  in  the  direction  of  the 
resultant.     Hence  such  forces  are  termed  forces  of  translation. 


10 


MECHANICS  OF   THE   GIRDER. 


But,  in  the  case  now  under  consideration,  we  have  two  forces 
in  one  plane  acting  at  two  points  in  a  rigid  body,  the  one  force 
at  one  point  tending  to  turn  the  body  about  the  other  point. 

I  say  two  forces,  for  it  is  manifest,  that,  at  the  point  which 
is  taken  as  the  centre  of  rotation,  there  must  be  a  resistance  to 
motion  equal  and  opposite  to  the  rotatory  or  tangential  force 
acting  at  the  other  point.  Such  a  system  of  two  parallel  forces 
acting  in  opposite  directions  is  called  a  couple,  and  the  perpen- 
dicular drawn  to  the  lines  of  action  of  the  forces  is  called  the 
arm  of  the  couple. 


FIG.  6. 


In  Fig.  6,  let  AOB  be  a  rigid  body,  beam,  or  bent  lever, 
whose  weight  is  not  now  to  be  taken  into  account ;  and  let 
AO  be  perpendicular  to  OB,  O  being  a  fixed  point  about  which 
the  applied  forces,  W^  and  H»  acting  at  right  angles  to  their 
respective  lever  arms,  tend  to  turn  the  beam.  Take  OB  =  /, 
and  OA  =  //,  where  /  and  h  represent  each  a  number  of  linear 
units.  Then  the  moment  of  the  force  Wl  is  —  WJ,  and  the 
moment  resulting  from  the  action  of  //i  is  HJi ;  giving  them 
different  signs  because  they  tend  to  turn  the  beam  in  opposite 
directions  about  the  point  O. 

If  these  two  moments  are  equal,  we  have 


-  WJ  =  o, 


(13) 


MOMENT  OF  A   FORCE.  II 

which  shows  that  there  is  no  rotation  about  the  point  O,  and 
that  the  three  forces  WI}  Hiy  and  the  resistance  to  translation 
offered  at  O,  are  in  equilibrium. 

At  the  fixed  point  O  are  developed  two  forces:  the  one, 
W2,  equal  to  Wlt  but  acting  in  the  opposite  direction,  OA  ; 
the  other,  H2,  equal  to  HJt  acting  at  O  in  the  direction  OB 
Now,  these  two  forces,  acting  at  the  same  point,  O,  must,  by 
Art.  3,  have  a  resultant  equal  to  the  diagonal  of  the  rect- 
angle of  which  W2  and  H2  are  the  adjacent  sides.  Therefore 
the  resultant  R  =  V IV22  +  H22,  and  the  tangent  of  the  angle 

between  the  line  of  R  and  the  line  OB  is  tan  0  =  ^  =  -, 

H2       / 

in  the  case  of  equilibrium,  or  when  the  two  forces  are  inversely 
proportional  to  their  lever  arms,  as  shown  in  equation  (13). 

This  resultant,  R,  is  a  force  of  translation,  and  may  be 
graphically  found  by  producing  the  lines  of  action  of  Wt  and 
HI  till  they  intersect  at  C;  then,  if  A  £7  represent  the  intensity 
of  //,,  and  BC  the  intensity  of  W»  we  have,  from  Art.  4,  R  = 
OC,  the  diagonal  of  the  rectangle. 

Otherwise,  graphically,  draw  BD  perpendicular  to  OC',  then, 
if  Wi  and  Ht  be  resolved,  each  into  one  component  along  OC 
and  one  at  right  angles  to  OCy  we  have 

Components  of  Hl   =  DO  and  BD, 
Components  of  W*  =  CD  and  DB. 
.'.     DO  +  CD  =  R  =  pressure  at  <9, 
BD  —  DB  =    o  =  rotatory  effect. 

If  we  suppose  the  rigid  body  extended  so  as  to  fill  the  space 
AOBC,  then  the  resultant  may  be  conceived  as  acting  at  any 
point  in  the  line  OC  without  altering  its  effect  of  translation 
on  the  whole  mass.  The  effect  within  the  body  will,  of  course, 
be  different  for  every  new  point  of  application.  With  this  we 
are  not  now  concerned. 


12  MECHANICS  OF   THE   GIRDER. 

We  conclude,  then,  that  if  two  forces  whose  lines  of  action 
are  in  the  same  plane  act  on  a  rigid  body,  and  if  from  any 
point  in  the  line  of  action  of  their  resultant,  perpendiculars  be 
drawn  to  the  lines  of  action  of  the  forces,  then  the  resistance 
at  the  point  chosen,  and  the  two  given  forces,  will  be  in  equi- 
librium, when  the  intensity  and  direction  of  the  resistance  are 
respectively  equal  and  opposite  to  those  of  the  resultant. 

This  conclusion  may  also  be  drawn  from  the  figure,  since 
two  lines  drawn  from  any  point  in  OC,  respectively  perpen- 
dicular to  the  lines  of  action  of  W^  and  H»  must  be  propor- 
tional to  /and  h,  and  therefore  equation  (13)  would  be  satisfied, 
whatever  be  the  angle  AOB. 

If  the  two  moments,  WJ  and  HJi,  are  not  equal,  let  us 
suppose  that  WJ  is  the  greater  by  reason  of  an  increment 
given  to  W»  so  that  /,  k,  and  H^  remain  unaltered.  Then  the 
resultant  of  the  forces  HT  and  W^  will  not  pass  through  the 
point  Oy  but  will  lie  somewhere  between  it  and  the  line  of 
action  of  the  augmented  force  Wv 

Suppose  CE  to  be  the  line  of  action  of  the  new  resultant 
RT,  and  draw  OE  perpendicular  to  CE ;  then  will  R^  X  OE 
represent  the  total  rotatory  effect  of  the  given  pressures 
Wi  and  Ht  with  respect  to  the  point  O,  and  we  shall  have, 
if  r  =  EO, 

-  WJ  +  HJi  =  -Rjr,  (14) 

where  —R^r  is  the  moment  of  a  couple,  equivalent  to  the  dif- 
ference or  algebraic  sum  of  the  moments  of  the  couples  whose 
arms  are  h  and  /. 

We  see,  then,  that  the  effect  of  one  couple  may  be  neutral- 
ized by  the  moment  of  another  couple  having  the  same  axis 
of  rotation  and  an  opposite  direction,  and  that  the  combined 
effort  of  two  couples  may  be  balanced  by  the  moment  of  a 
single  couple  having  the  same  centre  of  rotation. 

ii.  The   law  may  clearly  be   extended  to  any  number  of 


MOMENT  OF  A   FORCE.  13 

forces,  Plt  P2,  Pv  etc.,  acting  in  one  plane  to  turn  a  rigid  body 
about  a  fixed  point  in  that  plane,  or  about  a  fixed  axis  perpen- 
dicular to  that  plane.  Let  Plf  P2,  Pv  etc.,  be  the  lengths  of 
the  perpendiculars  drawn  from  the  fixed  centre  of  rotation  to 
the  respective  lines  of  action  of  Pu  P2,  Py  etc.  Let  R  be  the 
resultant  of  translation  of  all  the  forces,  and  r  the  length  of 
the  perpendicular  drawn  from  the  same  centre  to  the  line  of 
action  of  R  ;  then 

Rr  =  Pjt  +  P2p2  +  P,pz  +  etc.  =  ^Pp.          (15) 


The  algebraic  signs  of  the  terms  in  this  equation  will  depend 
upon  the  directions  in  which  the  forces  tend  to  turn  the  rigid 
body  ;  and  it  will  be  convenient  to  distinguish  moments  as 
positive  which  tend  to  turn  the  body  in  the  direction  taken 
by  the  hands  of  a  watch,  and  to  call  moments  having  the  oppo- 
site tendency  negative. 

For  equilibrium  we  must  have 

-$Pp  =  Rr  =  O.  (16) 

Equation  (16)  is  satisfied  either  when  R,  the  resultant  of  the 
given  forces,  becomes  zero,  or  when  r,  the  arm  of  the  resultant 
couple,  vanishes.  In  the  former  case  the  given  impressed  forces 
are  in  equilibrium  among  themselves  ;  in  the  latter  case,  if  R 
does  not  also  vanish,  it  is  equal  and  opposite  to  the  resistance 
offered  at  the  fixed  point. 

As  in  the  case  of  two  forces,  each  acting  tangentially  at 
one  extremity  of  its  lever  arm  to  cause  rotation  about  the 
fixed  point  common  to  the  other  extremities,  so  in  the  case 
of  many  forces  acting  in  one  plane  on  a  rigid  body,  and  tend- 
ing to  turn  it  about  a  fixed  point,  the  resistance  developed  at 
the  fixed  point  by  each  of  the  given  forces  will  be  equal  and 
opposite  to  the  given  force,  and  will  have  its  line  of  action 
parallel  to  that  of  the  given  force. 


MECHANICS  OF   THE   GIRDER. 


Hence  the  intensity  and  direction  of  the  resultant,  R,  may 
be  found  from  equations  (4),  (5),  (8),  and  (9),  as  in  the  case  of 
many  forces  acting  in  one  plane  at  a  common  point. 

12.  And,  having  found  Ry  equation  (15)  gives 


If,  then,  through  the  fixed  point  a  line  be  drawn  parallel  to 
the  line  of  action  of  R,  and  through  the  same  point  another 
line  be  drawn  at  right  angles  to  this  line  of  action,  and  if  on 
this  second  line  the  distance,  r,  be  laid  off  from  the  fixed  point, 
and  R,  both  in  magnitude  and  direction,  be  applied  at  the  outer 
extremity  of  r,  we  shall  have  a  graphical  representation  of  the 
resultant  couple  whose  moment  is  equivalent  to  the  combined 
action  of  all  the  given  forces. 

13.  The  direction  of  R  and  the  sign  of  Rr  will  show  on 
which  side  of  the  fixed  point  r  must  be  laid  off. 


Y' 


FIG.  7. 


EXAMPLE.  —  Let  ABCO  be  a  rigid  body  acted  on  by  five 
forces,  whose  lines  of  action  are  in  one  plane,  and  which  tend 
to  turn  the  body  about  the  fixed  point  O,  Fig.  7. 


MOMENT  OF  A   FORCE.  15 

Let  the  directions  of  the  forces  Pt,  P2,  Py  etc.,  and  their 
points  of  application,  be  as  indicated  in  the  figure ;  and  desig- 
nate the  angle  between  the  line  of  action  of  any  force  and  the 
axis  of  x  by  a,  and  the  distance  of  O  from  any  point  of  appli- 
cation by/.  Take 

Px  =     10  tons,     /,  =    4  feet,      a,  =  270°  or  —90°. 

P2  =  ioo  tons,     /2  =  12  feet,      a2  =  290°  or  —70°. 

P3  =    20  tons,      /3  =    8  feet,      «3  =  120°. 

P4  =    30  tons,      /4  =  10  feet,       «4  =  315°  or  —45°. 

P5  =  200  tons,     ps  =    6  feet,      as  =  180°. 

To  find  R,  let  these  forces  be  considered  as  acting  at  the 
point  O  in  direct  opposition  to  the  resistances  there  developed ; 
then,  by  equations  (4),  (5),  (8),  and  (9),  we  have 

X  =  10  cos  270°  4-  ioo  cos  290°  -f  20  cos  120°  +  30  cos  315° 

4-  200  cos  180°  =  — 10  cos  90°  -f-  ioo  cos (  —  70°)  —  20  cos  60° 
4-  3ocos(  —45°)  —   200  cos  o°  =   10  x  o  4-  ioo  x  0.34202 

—  20  X  0.5  -f-  30  X  0.70711  —  200  X  I  =  O  4-  34.202  —  10 

4-  21.2133  -  200  =  -154.5847. 

Y  =  losin  270°  -f-  ioo  sin  290°  -f-  20 sin  120°  4-  30 sin  3 15° 4  200 sin  1 80° 
=  10  x  — i  +  ioox  — 0.939694-20x0.86603  +  30  x  —0.70711 
4-  200  x  o  =  —10  —  93.969  4-  17.3206  —  21.2133  4-  o 
=  —107.8617. 


£=.  V(i54.5847)2  +  (-io7.86i7)2  =  188.496  tons. 

-'•<»"«• 


0  =  34°  54'  20"  ;  or,  since  X  and  Y  are  both  negative,  we 
must  have 

e  =  2  14°  54'  20", 

and  the  resultant  is  therefore  in  the  third  quadrant. 


1  6  MECHANICS  OF  THE   GIRDER. 

From  equation  (15), 

Rr=  2J%>=  +  10  x  4  -f  ioo  x  12  —  20  x  8  +  30  x  10  —  200  x  6 

=  40  -f-  1200  —  1  60  +  300  —  1200  =  +180  foot-  tons, 

•'•   r  -  =  °'95493  foot' 


Since  the  product  Rr  is  positive,  and  the  direction  of  the 
line  of  R,  when  drawn  through  the  fixed,  point,  is  into  the  third 
quadrant,  it  follows  that  r  must  be  laid  off  below  the  fixed  point 
on  the  perpendicular  to  the  line  of  R>  as  shown  by  OE  in 
the  figure  ;  E  being  supposed  rigidly  connected  with  the  solid 
AOCB. 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    1 7 


CHAPTER  III. 

MOMENTS    OF    THE    EXTERNAL    FORCES    APPLIED    TO    A    BEAM   OR 

GIRDER. 

SECTION  i. 

The  Semi-Beam,  or  Girder  fixed  at  One  End  and  free  at  the 

Other. 

• 

14.  We  can  now  find  expressions  for  the  moments  devel- 
oped in  any  section  of  a  beam  or  girder,  by  the  action  of  any 
forces  in  the  plane  of  the  beam,  in  whatsoever  manner 
applied 

Let  us  first  take  a  beam  fixed  at  one  end  and  free  at  the 
other,  or  a  semi-beam  as  it  is  called.  Let  EOAB,  Fig.  8, 
represent  a  beam  fixed  to  a  wall  along  the  line  AB  —  h.  Sup- 
pose the  weight  of  the  beam  to  be  w  pounds  for  every  unit  of 
its  length  /  =  AO.  Assume,  also,  that  the  length  b  =  DC  has 
an  additional  uniform  load  of  in/  pounds  per  linear  unit,  both 
w  and  w'  being  continuously  distributed  throughout  their  re- 
spective lengths.  Also  let  W  be  a  concentrated  weight  or 
pressure  at  the  distance  a'  —  BJ  from  the  fixed  end  of  the 
beam.  Let  BC  —  a  =  the  distance  from  the  wall  to  the  nearer 
end  of  the  uniform  load  ufb.  Let  P  be  any  pressure  acting  at 
any  point,  G,  with  any  inclination,  a,  to  the  arm  FG ;  and  call 
the  horizontal  distance  of  the  point  G  from  the  wall  a"  —  AK. 


18 


MECHANICS   OF  THE   GIRDER. 


Suppose  the  beam  horizontal,  and  all  the  applied  pressures, 
except  Pt  vertical.  Let  VS  be  any  vertical  section  of  the 
beam  at  the  distance  x  from  the  fixed  end  AB.  It  is  required 
to  find  the  moment  of  the  applied  forces  which  must  be  resisted 
by  the  internal  forces  of  the  beam  at  the  section  VS. 


W 


J          D 


a       B 


S 

FIG.  8. 


Manifestly  only  the  pressures  at  the  left  of  VS  affect  that 
section.  Taking  the  moments  of  these  sinister  pressures  about 
any  point,  F,  in  the  vertical  section  VS,  and  remembering  that 
downward  pressures  on  the  left  of  VS  give  negative  moments, 
we  have  the  following  equations  for  the  required  moment 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    19 


SEMI-GIRDER.     LENGTH  =  /.     (See  Fig.  8.) 


Load. 

Conditions. 

Force  left  of 
VS. 

Arm. 

Moments  about  F. 

w 

x  <a' 

W 

a'  -x 

M=-W(a'  -x). 

(18) 

w 

x  =  or  >  a' 

0 

M=o. 

(19) 

w 

x  •=.  o 

W 

a' 

M  =  —  Wa'. 

(20) 

w 

a'  =  1 

W 

1  —  X 

M  =  —  W(l  —  x). 

(21) 

w 

x  =  o,  a'  =  / 

W 

I 

M=  _  0V  (max.). 

(22) 

wl 

x<l 

w(l  -  x) 

W  -  x) 

M  =  —\w  (1  —  x)2. 

(23) 

wl 

X  =  1 

o 

M=o. 

(24) 

wl 

x  —  o 

wl 

V 

M  -  —\wlz  (max.). 

(25) 

w'b 

x>aznd<(a+t>) 

w'(a+b  -  x) 

\(a+b-x) 

M  =  —W  (a  +  b  —  x)2. 

(26) 

w'b 

a  =  o 

w1^  —  x) 

W  -  *) 

M  =  —\w'  (b  —  x}2. 

(27) 

w'b 

a  =  o,  b  =  I 

w'(l  -  x) 

i(/  -  x) 

M  =  -\w'  (1  —  x)2. 

(28) 

w'b 

x  =  or  <  a 

w'b 

\b  +  a  —  x 

M  =  —w'b  (\b  +  a  —  x). 

(29) 

w'b 

x  =  o,a=o,l>  =  t 

w'l 

i/ 

M  =  —\w'l2  (max.). 

(30) 

P 

x<a" 

Ps'ma 

P 

M  =  P  sin  ap. 

(31) 

i5«  In  applying  these  formulas  for  W  to  the  case  of  many 
equal  weights  placed  at  equal  intervals  along  the  beam,  we  may 
simplify  the  numerical  computations  by  first  summing  the  series 
resulting  from  assigning  to  a'  and  x  their  proper  values. 

Suppose  we  have  n  weights,  each  equal  to  W,  at  intervals 

of   -  feet  along  the  beam  ;  then 

The  moment  at  the  fixed  end  of  the  beam,  or  when  x  =  o, 
due  to  all  of  the  equal  weights  is,  by  summing  (20), 


M  =  -  WSJ  =  -  Wl(-  +  -  +  ^  +  .     .  -\ 
\n       n       n  nj 


mi* 

\n 

n  +  i 


-Wl 


(32) 


2O 


MECHANICS  OF  THE   GIRDER. 


The  moment  at  the  fixed  end  due  to  i,  2,  3,  ...  r,  of  these 
equal  weights,  first  in  order  is 


M  =  - 


(33) 


The  moment  at  the  fixed  end  due  to  the  remaining  (n  —  r) 
of  these  equal  weights  is 


n  n 


Wl(     a  >*       } 


(34) 


The  moment  at  the  interval  r  due  to  these  remaining  (n  —  r) 
equal  weights  is 


M=  - 


n      n      n 


(35) 


EXAMPLE  I.  —  A  semi-girder  projects  50  feet,  and  is  loaded 
at  intervals  of  10  feet  with  a  weight  of  10  tons  ;  required  the 
moment  at  the  fixed  end  due  to  the  5  equal  weights. 

From  (32), 


M=  —Wl 


10  x  50  x  -  = 

2 


-1500  foot-tons. 


EXAMPLE  2.  —  The  same  conditions  continuing,  required  the 
moment  at  the  fixed  end  due  to  the  first  3  of  the  weights. 
From  (33), 


M  =  -  Wl 


rr 


-10  X  50  x 


-600  foot-tons. 


2X5 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   21 

EXAMPLE  3.  —  With  same  conditions  of  beam  and  load,  re- 
quired the  moment  due,  at  the  fixed  end,  from  the  remaining 
2  weights. 

From  (34), 


M  =  - 


*  -  3(3_±jO\  _        00  foot-tons. 
2  x  5    / 


EXAMPLE  4.  —  At  10  feet  from  the  fixed  end  of  the  beam, 
what  is  the  moment  due  to  the  4  weights  beyond  ? 
From  (35), 

M  =  -  Wl^  ~  T  +  l)(5  ~  T)  =  -looo  foot-tons. 
2  X  5 

EXAMPLE  5.  —  If  the  given  semi-beam  weighs  0.8  ton  to 
the  linear  foot,  what  is  the  moment  at  its  centre  and  at  its 
fixed  end  ? 

From  (23),  if  x  =  £/, 

M=  —\w(\lY  —  \  X  0.8  X  so2  =  —250  foot-  tons. 
From  (25), 

M=  —  %  X  0^8  x  502  =  —looo  foot-  tons. 

EXAMPLE  6.  —  Suppose  the  same  beam  to  be  covered  with 
the  uniform  load  0.6  ton  for  the  space  of  15  feet,  beginning  25 
feet  from  the  fixed  end  ;  required  the  moment  due  to  this  load 
at  30  feet  from  the  fixed  end. 

Here 

a/  =  0.6,        b  =  15,        a  =  25,        x  =  30. 
From  (26), 

M=  —  |  x  0.6(25  +  15  —  30)*  =  —30  foot-  tons. 


22  MECHANICS  OF   THE    GIRDER. 

EXAMPLE  7.  — If  the  load  0.6  ton  per  foot  covers  the  first 
35  feet  of  the  beam,  and  the  moment  at  10  feet  is  required,  we 
have  b  =  35,  a  —  o,  x  =  10;  and,  from  (27), 

M  —  —  \  X  0.6(35  —  I0)2  =  — 187-5  foot-tons. 

EXAMPLE  8.  —  If  the  load  0.6  ton  per  foot  covers  the  entire 
beam,  the  moment  at  the  centre  is,  from  (28), 

M  =  —  |  X  0.6  X  (50  —  25)2  =  —187.5  foot-tons, 
and  at  the  fixed  end 

M—  —  |  X  0.6  X  5o2  =  —  750  foot-tons. 

EXAMPLE  9.  —  If  the  uniform  load  0.6  ton  covers  40  feet 
of  the  beam,  beginning  at  the  free  end,  then  the  moment  at  5 
feet  from  the  fixed  end  is,  from  (29), 

M=  —0.6  X  4o(J  X  40  +  10  —  5)  =  —600  foot-tons. 

EXAMPLE  10.  —  If  the  force  P,  Fig.  8,  is  4  tons,  and  its  line 
of  action  makes  an  angle  of  30°  with  the  line  GF  —  P  =  20 
feet,  then  the  moment  due  to  P  at  the  point  Fis,  from  (31), 

M  —  4  X  0.5  X  20  =  40  foot-tons. 

1 6.  If  there  are  several  concentrated  weights,  W»  W2,  Wy 
etc.,  or  pressures,  Plt  P2,  Py  etc.,  or  detached  uniform  loads, 
bjWTrt  b2w21  b^w^  etc.,  at  different  points  on  the  left  of  the 
section  VS,  we  must  evidently  sum  the  moments  due  to  the 
separate  pressures  for  the  total  moment. 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   2$ 

Thus  we  may  write 

Mx  =  -^W(a'  -  x)  -  \w(l-  x)2  -  %uf(a  +  b  -  x)2 

+  a  —  x)b  +  S/'sina/,     (36) 


where  Mx  is  the  moment,  with  reference  to  any  point  of  any 
vertical  section  of  a  semi-beam,  due  to  all  the  forces  applied  to 
the  beam  between  its  free  end  and  the  given  vertical  section. 

It  should  be  observed,  that,  for  all  pressures  whose  lines  of 
action  are  vertical,  the  moments  will  be  the  same,  whatever 
point  of  reference  is  taken  in  the  vertical  section  VS  \  for  such 
pressures  have  no  horizontal  component. 


SECTION  2. 

17.  We  next  take  a  beam  or  girder,  horizontal,  supported  at 
its  ends,  and  loaded  in  any  manner  whatsoever.  Such  a  girder 
is  also  said  to  have  its  ends  free;  since  they  simply  rest  upon 
two  level  supports,  and  are  fixed  in  no  other  manner. 


x 


FIG.  9. 

Let  the  beam  OABE,  Fig.  9,  be  supported  at  the  two  points 
O  and  A,  and  be  subjected  to  the  following  pressures  :  — 


24  MECHANICS  OF   THE   GIRDER. 

w    =  weight  of  beam  per  linear  unit. 

w'  =  uniform  load  per  linear  unit  of  the  length  b  =  CD. 

W  =  a   concentrated   weight   or  vertical  pressure   at  any 

point,  K. 

P    =  any  force  at  distance  a"  =  OF  from  O. 
Fx  =  vertical  re-action  of  the  left  support. 
V*  =  vertical  re-action  of  the  right  support. 
/     =  length  of  girder. 
h    =  height  of  girder. 
a    =  OD  =  the  horizontal  distance  from  the  origin  0  to 

the  nearer  end  of  the  uniform  load  bud. 
a'    =  the  horizontal  distance  from  O  to  the  weight  W. 
.x    =.  the  horizontal  distance  of  any  vertical  section,    VS, 

from  O,  the  origin  of  co-ordinates. 
•y    =,  the  vertical  distance  of  any  point  in  the  section  VS 

from  the  horizontal  line  AO. 


The  vertical  section  VS  is  made  by  any  plane  cutting  the 
'beam  perpendicular  to  the  line  AO. 

It  is  required  to  find  the  moment  generated  at  any  vertical 
.section,  VS,  by  the  action  of  each  of  the  given  pressures. 

Since  at  any  given  cross-section,  there  can  be  but  one  mo- 
.ment  due  to  the  given  simultaneous  pressures,  it  follows  that 
we  may  determine  this  moment,  either  by  using  the  pressures 
applied  upon  the  left  side  of  the  given  section,  or  by  using  the 
applied  pressures  on  the  right  side  of  the  same  section. 

In  the  following  table  we  use  the  pressures  that  act  on  the 
left  of  the  section  VS ;  and  consequently  downward  pressures 
give  negative  moments,  and  upward  pressures  give  positive 
moments,  in  accordance  with  our  previous  notation. 

18.  The  sum  of  the  re-actions  VT  and  V2  for  the  simple 
girder  with  free  ends  is  equal  to  the  total  weight  of  the 
girder  and  its  load. 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    2$ 

The  resistances  Vt  and  V2  due  to  any  concentrated  weight, 
W,  are,  since  there  can  be  but  one  moment  for  the  vertical 
section  through  Wy  inversely  proportional  to  the  horizontal 
distances  of  W  from  the  points  of  support  ;  and  we  have,  from 
equation  (13), 

M     =  VJ  =  V2(l-  of), 

...         Fi      =  F2^=  W-  F2,  (37) 


K2     =  WaT  (38) 

Vt     =  W1^-.  (39) 


Or,  by  proportion, 

F,  :  F2  : :  /  -  a'  :  J, 


Similarly,  for  the  uniform  load  bit/,  the  re-actions  Fx  and  Z^ 
will  be  inversely  proportional  to  the  distances  of  the  centre  of 
gravity  of  the  uniform  load  from  the  points  of  support. 

19.  In  the  following  table  we  have, — 

First  column,  load  whose  moment  is  sought. 

Second  column,  re-action  at  left  support,  giving  -\-M. 

Third  column,  conditions  of  load  and  plane  VS. 

Fourth  column,  part  of  load  on  left  of   VS,  giving  — M.  - 

Fifth  column,  arm  of    Vv 

Sixth  column,  arm  of  load  on  left  of   VS. 


26 


MECHANICS  OF  THE   GIRDER. 


S      S      'ft 


II       II 


d  d 

II    II 


v§      >»      >      vg 


:  ^ 

*  x 


i 


v« 

°     i 


H          H 


^  5Q  O    ^    *        «  « 

J  II     II     II      V          II 

J        H  *   *    *     *        H 


»5R   .9       .£ 


* 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    2J 

20.  Moments  due  Uniform  Discontinuous  Load  on  any 
Part  of  the  Beam.  —  Let  rt  —  r2  denote  the  number  of  equal 
weights,  W,  at  equal  intervals,  c,  between  any  two  consecutive 
weights  on  the  whole  or  any  part  of  the  girder.  We  may 
shorten  the  numerical  computation  of  moments,  as  in  case  of 
the  semi-girder,  by  first  summing  the  series  that  arises  in  the 
expression  for  M. 

For  this  purpose  let  r  —  r2  =  the  number  of  equal  weights, 
W,  on  the  length  x\  (r2  +  i)c  =  the  distance  from  the  left  end 
of  the  beam  to  the  nearest  weight.  If  this  distance  is  less  than 
c,  that  is,  not  a  full  interval,  it  follows  that  r2  will  be  a  negative 
proper  fraction.  Now  (r,  —  r2)  —  (r  —  r2)  =  r,  —  r  =  the  num- 
ber of  equal  weights  between  the  point  x  and  the  right  end  of 
the  beam.  The  three  differences,  rx  —  r2,  r  —  r2,  and  rl  —  r, 
must  be  integers,  since  each  denotes  a  number  of  equal  weights. 
If  one  of  the  three  quantities  r,  r»  r2,  is  not  an  integer,  neither 
of  the  other  two  is  an  integer,  and  the  decimal  part  of  each  is 
the  same,  except  that,  when  r2  is  negative,  its  value  is  less  by 
unity  than  the  common  decimal  part  of  r  and  r,. 

Let  us  first  find  the  moment  due  r—  r2  equal  weights,  W, 
at  any  point,  x,  between  the  last  weight  and  right-hand  end  of 
the  girder.  We  use  equation  (43),  giving  to  af  the  successive 
values  c(r2  +  i),  c(r2  +  2),  c(r2  +  3),  .  .  .  cr,  and  taking  the 
sum ;  thus, 

^a'  =  c\_(rz  +  i)  4-  (ra  +  2)  +  (ra  +  3)  +  -  -  •  r] 

=  &(r-r2)(r  +  r2  +  i), 

...    Mx=  ^(r  Ir2)(r  +  r2  4-  0 (/-*),          (60) 
where  x  cannot  be  less  than  cr. 

EXAMPLE  i.  —  Length  of  beam  =  /  =  100  feet  =  ioc,r  =.  6J, 
r2  =  2\ ;  what  is  the  moment  at  the  fourth  weight,  W=8  tons, 
due  the  4  weights  =  32  tons  ? 


28  MECHANICS  OF  THE   GIRDER. 

Here  x  =  re  =  65, 

.-.    Mr  =  —  —  —  x  4  X  10  x  35  =  16  X  35  =  560  foot-tons. 
2  x  100 

If  x  =  (r  +  i)c  =  75, 

.'.  Mr  +  I  =  1  6  X  25  s=  400  foot-tons. 
If  *  =  (r  +  2)*  =  85, 

.*.  Mr  +  *  =  16  X  15  =  240  foot-tons. 
If  *  =  (r  +  3)*  =  95, 

.*.    -^-  +  3  =  16  x     5  =    80  foot-tons. 

This  shows  a  uniform  decrease  of  moment  for  each  interval 
beyond  the  given  load. 

Equation  (40)  gives  for  a  single  weight,  W,  applied  at  any 
point,  a',  the  moment  at  any  distance,  x,  between  the  weight 
and  the  left  end  of  the  beam.  By  giving  to  a'  the  successive 
values  c(r  +  i),  c(r  +  2),  c(r  +  3),  ...  crlt  and  summing  for 
aT  and  of,  we  find 

-  r,  -  r, 


(61) 


which  is  the  moment  at  any  point,  x,  between  the  left  end  of 
the  girder  and  the  nearest  weight,  which  is  at  the  (r  -\-  i)ih 
point  of  division  ;  the  number  of  weights  being  r,  —  rt  and  x 
not  being  greater  than  c(r  +  i). 


MOMENTS   OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    2Q 

EXAMPLE  2.  —  Let  /=  100  feet  =  loc,  W=  8  tons,  rr  =  6J, 
r  =  2j;  what  is  the  moment  at  the  first  weight  due  the  rt  —  r 
=  4  equal  weights  ? 

Here  x  —  (r  +  i)c  =  35, 

o 

(2  X  4  X  100—  10  X  4  X  10)35  =  560  foot-to 


2  X  100 
x  =  re  =  25, 

.*.    J/r       =  1  6  X  25  =  400  foot-tons. 
x  =  (r  —  i)c  =  15, 

.*.    Mr_i  =  16  X  15  =  240  foot-tons. 
#  =  (r  -  2)<r  =     5, 

.'.    Mr_2  =  16  x    5  =    80  foot-tons. 

This  shows  a  uniform  decrease  of  moment  for  each  interval 
before  the  given  load.  These  moments  are  the  same  as  those 
of  example  i,  as  they  should  be,  since  the  same  load  is  symmet- 
rically placed  on  the  beam  in  both  cases. 

If  we  add  equations  (60)  and  (61),  calling  the  sum  Mx  still, 
we  shall  have 


cl(r  -  ra)(r  +  ra  +  i),    (62) 


which  is  the  moment  at  any  point,  x>  of  the  beam  due  rI  —  r2 
equal  weights,  W,  placed  at  equal  and  consecutive  intervals,  c, 
over  the  whole  or  any  part  of  its  length. 


3O  MECHANICS  OF  THE   GIRDER. 

Here  r  cannot  be  less  than  r2  nor  greater  than  rn  and  x  lies 
between  re  and  (r  +  i)c  for  the  loaded  part  of  the  beam,  but 
may  have  any  value  between  o  and  r2c  where  r  —  r2,  and  any 
value  between  r,c  and  /  where  r  =  TV 

EXAMPLE  3.  —  Let  /  =  100  feet  =  10  c,  W  =  8  tons,  r2  = 
2j,  r,  =  6J  ;  what  is  the  moment  at  the  fourth  weight  ?  Here 
x  =  r^c  =  65  feet,  and  (62)  gives, 

If  r  =  riy 

Mri  =  —  |  (o  —  10x4X10)65-1-10x100x4X101  =  560  foot-tons. 
200  \  ) 

Or,  if  r  =  rl  —  i, 

g        f  -v 

Mri  =   -      j  (2X  i  X  100  —  10X4  X  10)65  +  10  X  100X3X9  t 

=  560  foot-  tons. 

What  is  the  moment  one  interval  beyond  the  last  weight  ? 
Here  x  =  (r,  -f-  i)c  =  75,  and  r  =  r,  =  6J,  in  (62)  ; 

8    C  *  ) 

.*.     J/ri  +  I  =  --  \(o  —  10x4X10)75  +  10x100X4X101 

200  (  j 

=  400  foot-tons. 

If  n  =  the  whole  number  of  intervals  in  the  girder's  length, 
we  have  c  =  -,  and  (62)  becomes 


Mx  =          [2»(rs  -  r)-(rs  -  ra)(rx  +  r2  +  i)] 

/ 

,      (63) 


from  which  we  may  find  the  simultaneous  moments  at  all  points 
throughout  the  girder  due  to  any  uniform  discontinuous  par- 
tial or  full  load. 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   31 

EXAMPLE  4.  —  Let  a  uniform  load  of  4  weights,  each  =  W 
=  8  tons,  spaced  c  =  10  feet  from  weight  to  weight,  come  upon 
a  girder  loofeet  long,  and  move  forward  to  the  centre ;  required 
the  simultaneous  moments  throughout  the  girder  as  the  fore- 
most end  of  the  load  passes  the  points^  =  5,  15,  25,  35,  etc., 
n  =  10. 

Owing  to  the  important  applications  of  this  formula  which 
are  to  follow,  we  add  the  complete  solution  of  this  problem, 
and  may  remark  that  by  giving  to  x  the  values  10,  20,  30,  40, 
etc.,  we  can  find  the  simultaneous  moments  at  the  full  intervals 
as  the  foremost  end  of  the  load  passes  the  successive  points  of 
division.  Or  we  may  give  x  any  value  we  please  between  o  and 
/,  and  so  suit  the  equal  or  unequal  panel  lengths  of  any  girder. 
As  the  load,  now  central,  passes  off  to  the  right,  it  is  evident 
these  moments  will  be  reversed. 


UNIVERSITY 


MECHANICS   OF   THE   GIRDER. 


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MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    33 

In  the  above  table  all  moments  included  by  the  same  brace 
are  simultaneous,  and  due,  at  their  respective  points,  to  all  the 
weights  on  the  girder,  as  indicated  by  the  first  column,  rt  —  r2. 

Only  the  first  moment  in  any  horizontal  line  is  computed 
by  the  formula  in  that  line  ;  the  remaining  moments  in  any  line 
being  found  by  the  simple  variation  of  x,  using  only  the  term 
containing  x.  In  this  example  the  constant  difference  to  be 
added  to  the  first  moment  in  any  horizontal  line  of  moments  is. 
four  times  the  quantity  in  the  parenthesis  for  the  given  line. 

21.  Let  the  length,  /,  of  the  girder  be  divided  into  n  equal 
intervals,  c,  so  that  there  are  n  —  i  points  of  division  ;  then,  if 
a  weight,  IV,  be  applied  at  each  point  of  division  beginning  at 
the  left,  we  may  find  the  moment  at  the  foremost  end  of  this 

7 

advancing  load,  from  equation  (60),  by  putting  r2  =.  o,  x  =  re  =  —.. 

n 
Thus, 


which  is  the  moment  at  the  foremost  end  of  a  uniform  discon- 
tinuous load  when  that  end  passes  the  rth  point  of  division,  and 
r  equal  weights  have  come  on. 

EXAMPLE.  —  Let  /  =  100  feet,  n  =  10,  W  =  8  tons  ;  what 
is  the  moment  at  each  point  of  division  as  the  foremost  end 
of  this  load  passes  it  ?  Using  (64), 


If  r  =  i,  M, 

2,  M2 

3>M3 

4,  ^4 

6,  M6 

7,M7 

9>Mg 

= 

i 

2 

3 

4 

5 
6 

7 
8 

9 

X 
X 
X 
X 
X 
X 
X 
X 
X 

2 

3 

4 

5 
6 

7 
8 

9 
10 

X 
X 
X 
X 
X 
X 
X 
X 
X 

9 

8 

7 
6 

5 
4 
3 

2 

I 

X 
X 
X 
X 
X 
X 
X 
X 
X 

4 
4 
4 
4 
4 
4 
4 
4 
4 

- 

72 
192 

336 
480 
600 
672 
672 

576 
360 

foot-tons, 
foot-  tons, 
foot-tons, 
foot-tons, 
foot-tons, 
foot-tons, 
foot-tons, 
foot-tons, 
foot-tons. 

34  MECHANICS  OF   THE   GIRDER. 


22.  From  equation  (63),  by  putting  r2  =  o,  rl  =  n  —  i,  and 


r/ 

=  ?r  —  —  ,  we  derive 
n 


(65) 


-which  is  the  moment  at  the  rth  weight  due  «  —  i  equal  weights, 
W,  placed  at  equal  intervals,  -,  throughout  the  girder. 


EXAMPLE.  —  Uniform   discontinuous   load  ;     W  =  8  tons, 
/  =  100  feet,  n  —  10. 

If  r  =  i,  MI  =  40  x  9  X  i  =  360  foot-tons. 

2,  M2  =  40  X  8  X  2  =  640  foot-tons. 

3,  J/3  =  40  X  7X3=  840  foot-tons. 

4,  J/4  =  40  x  6  X  4  =  960  foot- tons. 
5,^  =  40x5x5  =  1000  foot- tons. 

And  these  moments  are  to  be  reversed  for  the  other  half- 
span. 

23.  Suppose  that  the  first  and  last  intervals  into  which  the 

beam  is  divided  are  each  =  \c  =  — ,  while  every  other  is  =  c, 

2n 

and  that  we  wish  to  find  the  moment  at  the  foremost  end  of  a 
uniform  load  of  equal  intervals,  c,  as  that  end  passes  each  point 
of  division  of  the  beam. 

For  this  object  we  employ  equation  (60),  makings  =  re  = 

— ,  and  r2  =  — J,  and  have 


n " r)- (66) 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    35 


EXAMPLE.  —  Let  /  = 

:  ioo  feet 

,  w=s 

tons,  n  •=.  10. 

For  r  =  0.5,  Mt 

=  4 

x 

I 

X 

9i  =    38 

foot-tons. 

i-5,  M2 

=  4 

X 

4 

X 

8*  =  136 

foot-tons. 

2.5,  M3 

=  4 

X 

9 

X 

7i  =  270 

foot-tons. 

7.S,    M. 

\J   <J  s           4 

=  4 

X 

16 

X 

6^  =  416 

foot-tons. 

4.5,  MS 

=  4 

X 

25 

X 

5i  =  55° 

foot-tons. 

5-5>  ^ 

=  4 

x 

36 

X 

4i  =  648 

foot-tons. 

6.5,  J/7 

=  4 

X 

49 

X 

3j  =  686 

foot-tons. 

7-5,  ^s 

=  4 

X 

64 

X 

2^   =    640 

foot-tons. 

8.5,  J/9 

=  4 

X 

81 

X 

ij   =    486 

foot-tons. 

9-5>  Mw 

=  4 

X 

IOO 

X 

^   =    200 

foot-tons. 

24.  From  equation  (63),  by  making  r2  =  —  J,  rt  =  n  — 


and  x  =  —  ,  we  also  obtain 
n 


(67) 


which  is  the  moment  at  any  weight  due  n  equal  weights,  Wf 
applied  at  equal  intervals,  -,  except  the  interval  at  each  end, 

which  is  =  — .     Here  r  takes  the  values  },  f ,  f ,  |,  .  .  .  2n  ""  l' 

21 1  2 

and  n  denotes  the  whole  number  of  full  intervals  in  the  length, 
/,  which  in  this  case  is  also  the  whole  number  of  weights. 

EXAMPLE.  —  Let,  as  before,  /  =  ioo  feet,  n  =  10,  W=8 
tons ;  then 

For  r  =  0.5,  M!  =  10(1  x  19  +  i)  =     200  foot-tons. 

1.5,  M2  =  10(3  x  17  -h  i)  =    520  foot-tons. 

2.5,  M3  =  10(5  X  15  +  i)  =     760  foot-tons. 

3.5,  M4  =  10(7  x  13  +  1)  =     920  foot-tons. 

4.5,  M5  =  10(9  x  ii  4-  i)  =  I0°°  foot-tons. 

The  same  to  be  reversed  for  the  other  half. 


36  MECHANICS  OF   THE   GIRDER. 

25.  Difference  of  Simultaneous  Moments  at  Consecu- 
tive Points  of  Division.  —  By  making  r2  ~  o,  and  x  = 
(r  +  i)ct  in  equation  (60),  we  have 


2H2 


«-r-  i),      (68) 


which  is  the  moment  one  interval,  c,  beyond  the  foremost  end 
of  a  uniform  load  consisting  of  r  equal  weights,  Wy  at  the  dis- 
tances c,  2c,  3^,  ...  re,  respectively,  from  the  left  end  of  the 
beam. 

Subtracting  (64)  from  (68),  we  have 


=  Mr  +  I  -Mr  =  -r(r  +  i),  (69) 

272 


which  is  the  difference  between  the  simultaneous  moments  at 
any  two  consecutive  points  of  division  on  the  unloaded  end  of 
a  beam,  which  has  r  equal  weights  at  full  intervals  on  the  other 
end. 

For  finding  the  difference  of  simultaneous  moments  at  con- 
secutive points  of  division  on  the  loaded  part  of  the  beam,  we 
use  (62),  making  r2  =  o,  and  x  —  re,  (r  +  i)c,  (r  -\-  2)c,  etc.,  in 
succession.  Thus, 


+  clr(r  +  i 
=—  J2(rI-r)»-r1(r1  +  i)l,         (70) 


which  is  the  first  order  of  differences  for  the  loaded  part  or 
end  of  the  beam  :  and  &M  is  an  increasing  function  of  r,  for 
a  given  value  of  r,  and  will  be  greatest  when  r,  is  greatest  ; 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   37 

that  is,  when  r,  =  n  —  I,  its  limit.  But  at  this  limit  of  rx  the 
girder  is  loaded,  and  the  positive  differences  on  the  left  half 
will  equal  the  negative  differences  for  the  corresponding  inter- 
vals on  the  right  half  of  the  girder.  Putting  n  —  i  for  rI  in 
(70),  we  have 

(71) 


which  gives  the  difference  of  simultaneous  moments  for  each 
interval,  c,  of  the  beam  fully  loaded  with  n  —  I  weights,  W, 
applied  at  equal  and  all  full  intervals,  c,  or  with  n  weights,  W, 
when  each  end  interval  =•  \c. 

Subtract  (69),  which  is  negative,  from  (71),  whose  positive 
values  in  one  half-span  are  equal  to  its  corresponding  negative 
values  in  the  other  half-span,  and  the  remainder  is 


which  is  positive,  since  n  >  r,  and  both  n  and  r  are  integers. 

Therefore  the  greatest  negative  difference  computed  by 
(71)  for  any  interval  is  numerically  less  than  the  difference 
computed  by  (69)  for  the  same  interval  in  the  second  half- 
span;  that  is,  both  half-spans,  if  the  uniform  load  is  to  travel 
either  way.  Consequently  we  use  (69)  in  finding  the  greatest 
difference  of  simultaneous  moments  for  any  interval  due  a  uni- 
form discontinuous  moving  load. 

It  may  be  observed  here  that  (69),  for  the  unloaded  end  of 
the  beam,  gives  a  constant  first  difference,  while  (70),  for  the 
loaded  end,  gives  a  first  difference  which  is  not  constant.  By 
putting  r  -f-  i  for  r  in  (70),  and  subtracting  (70)  from  the  re- 
sulting equation,  we  find  the  second  difference, 

A(A^f)  =  -Wf,  (72) 

which  is  constant  and  negative,  and  may  be  conveniently  em- 
ployed in  some  computations. 


MECHANICS  OF   THE    GIRDER. 


EXAMPLE  i.  —  Let  a  girder  of  10  panels,  each  10  feet,  be 
laden  with  a  permanent  load  of  4  tons  at  each  panel  point,  and 
a  discontinuous  uniform  rolling  load  of  8  tons  to  be  applied  at 
the  same  points  as  the  load  advances ;  required  the  greatest 
moments  at  these  panel  points,  and  the  greatest  difference  of 
simultaneous  moments  at  any  two  consecutive  panel  points,  due 
to  both  these  loads. 

The  greatest  moments  will  occur  when  both  loads  cover  the 
beam.  We  have,  then,  in  equation  (65),  W=  12,  /=  100,  n  = 

10,  —  ==•  60,  and  r  =  i,  2,  3,  etc.,  in  succession,  for  the  greatest 
2n 

moments. 

The  difference  of  moments  at  consecutive  panel  points  due 
dead  load  is  to  be  computed  by  (71),  making  W  =  4,  c  =  10, 

n  =  10,  —  =  20,  and  r  =  o,  i,  2,  3,  4,  etc.,  in  succession. 

2 

And  the  greatest  difference  of  simultaneous  moments  for 
each  interval  due  live  load  is  found  by  using  equation  (69), 

We 

when  W  =  8,  c  =  10,  n  =  10,  —  =  4,  and  r=  i,  2,  3,  4,  etc., 

2n 

in  succession. 


COMPUTATION  FOR  GREATEST  MOMENTS  AND  DIFFERENCES. 


No.  of  the  Panel  Point,  r. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Greatest  moments 

=  60(10  —  r)r 

0 

540 

960 

1260 

1440 

1500 

1440 

1260 

960 

S4o 

Differences,  dead  load 

=  20(9  —  2r) 

1  80 

140 

100 

60 

20 

—  20 

—60 

—  IOO 

—140 

—  180 

Greatest  differences,  live 

load        =  —  4/-(r  +  i) 

0 

—8 

—24 

-48 

—80 

—  120 

—  168 

—224 

—288 

-360 

Total  differences  for  both 

loads 

1  80 

TO2 

76 

12 

—60 

IAO 

—  228 

'32J. 

—428 

—  54° 

Differences,  load  moving  ( 

X  3* 

yu 

—60 

AT° 
—I40 

—  228 

O^T- 

—324 

-428 

—54° 

either  way                       ( 

540 

428 

324 

228 

140 

60 

MOMENTS  OF  FORCES  APPLIED    TO   BEAM  OR   GIRDER.   39 

If  in  equation  (66),  instead  of  the  factor  (/  —  re],  we  write 
[/  —  (r  +  i)c],  and  then  subtract  (66)  from  the  resulting  equa- 
tion, we  shall  have 

AJ/=  --(r-M)2,  (73) 


which  gives  the  greatest  difference  of  simultaneous  moments 
at  any  two  consecutive  points  of  division,  due  live  load  advan- 
cing by  equal  panel  weights,  when  the  two  extreme  panels  have 
each  but  half  the  length  of  every  intervening  panel.  Here  ob- 

serve that  r  takes  the  successive  values  —  J,  J,  f  ,  f  ,  .  .  .  — 

and  that  c  =  —  for  the  two  extreme  panels,  but  c  =  -  for  all 
2H  n 

others. 

EXAMPLE  2.  —  Given  the  same  loads  and  length  of  girder 
as  in  example  I,  but  the  panel  points  being  now  at  the  distances 
5,  15,  25,  35,  etc.,  from  either  end  ;  required  the  greatest  mo- 
ment at  each  of  these  points,  and  the  greatest  difference  of 
simultaneous  moments  at  any  two  consecutive  panel  points,  for 
both  live  and  dead  loads. 

Equation  (67)  gives  the  greatest  moments  if  we  make  /  = 

Wl 
100,  n  =  10,  W  =  12,  —  =  15,  and  r  =  |,  f,  f,  |,  etc.,  in  suc- 

oH 

cession. 

The  differences  for  dead  load  are  computed  from  (71)  by 
putting  W  =  4,  c  =  5  in  two-end  panels,  c  =  10  in  all  others, 

n  —  10,  and  r  =  —  J,  J,  f,  f  ,  etc.,  —  -  =  10  or  20. 

The  greatest  differences  for  live  load  are  found  from  (73), 
where  W  =  8,  c  =  5  or  10,  n  =  10,  and  r  =  —  J,  J,  f,  f,  etc., 


2n 


4o 


MECHANICS  OF   THE    GIRDER. 


COMPUTATION   FOR   GREATEST   MOMENTS  AND  GREATEST   SIMULTANEOUS 
DIFFERENCES  FOR  EACH  INTERVAL. 


r. 

-* 

i 

1 

I 

I 

f 

* 

¥ 

¥ 

¥ 

¥ 

Greatest  moments, 

U/-7I                                  i 

£K-'H 

300 

780 

1140 

1380 

1500 

1500 

1380 

1140 

780 

300 

Differences,  dead  load, 

^£(«_i_2r) 

100 

160 

120 

80 

40 

0 

—40 

—80 

120 

—  160 

—  TOO 

Differences,  live  load, 

~  (r  +  %)  2 

0 

—4 

—  16 

-36 

-64 

—  100 

—144 

—  196 

-256 

—324 

2OO 

Total  differences  .     . 

100 

156 

104 

44 

—24 

100 

-184 

-276 

-376 

-484 

—300 

Differences  to  be  used  j 

300 

484 

376 

276 

—24 

184 

—  100 
100 

-184 

24 

-276 

-376 

-484 

—  300 

26.  To  determine  the  Point  in  any  Girder  simply  sup- 
ported at  its  Two  Ends,  and  carrying  any  Partial  or  Com- 
plete Uniform  Discontinuous  Load,  where  the  Moment  due 
that  Load  is  Greatest.  —  The  required  greatest  moment  will 
occur  at  a  point  within  the  loaded  part  of  the  girder,  since  for 
any  partial  load  the  simultaneous  moments  decrease  from  either 
end  of  the  load  to  the  corresponding  end  of  the  girder. 

If,  therefore,  we  put  x  =  —  in  (63),  and  call  the  result  Mr, 

n 

then  in  Mr  thus  found   put  (r  +  i)  for  r,  giving  Mr  +  TJ  and 
equate  &Mr  =  Mr  +  x  —  Mr  to  zero,  we  shall  find 


r  =  r,  — 


(rt  - 


2H 


(74) 


and  the  panel  point  of  greatest  moment  lies  between  re  and 
(r  +  i)c,  except  when  re  and  (r  +  i)c  are  panel  points. 

Let  us  verify  this  statement  by  referring  to  example  4, 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    41 


article  20.     Taking  rlt  r2,  r,  and  the  greatest  moment,  from 
that  example,  we  compute  r  by  (74),  and  write  as  below :  — 


ft. 

r2. 

r. 

Mmax. 

r  by  (74). 

6.5 

2-5 

4-5  or  5-5 

640 

4-5 

5-5 

i-5 

4-5 

624 

3-9 

4-5 

°-5 

3-5 

544 

3-3 

3-5 

-°-5 

3-5 

416 

2.7 

2-5 

-o-5 

2-5 

270 

2.05 

i-5 

-°-5 

t-5 

136 

i-3 

0.5 

-°-5 

°-5 

38 

045 

r  I 
27.  If  in  equation  (63)  we  make  r  =  r,,  x  —  — ,  r2  =  r,  —  e, 

e  being  the  number  of  equal  weights  on  the  beam,  we  shall  find, 
after  putting  &Mfi  =  Mfi  +  x  —  Mfi  =  o, 

r.-2i±p*  (75) 

But  when  the  advancing  load  reaches  back  to  the  left  end 
of  the  girder,  we  may  not  know  how  many  weights  will  give  a 
maximum  moment  at  the  foremost  weight.  In  that  case  we 
deduce  AJ/r  =  Mr  +  x  —  Mr  —  o  from  (64),  and  find 

r  =  r,  =  \(n  -  i)  (76) 

for  a  girder  of  equal  panels  to  receive  an  advancing  load  of 
equal  weights  applied  at  successive  panel  points. 

And  for  a  girder  each  of  whose  two  extreme  panels  is  one- 
half  of  any  other,  the  advancing  load  to  be  applied  at  panel 
points,  we  derive  Aj/r  —  Mr  +  x  —  MT  —  o,  from  (66),  and  get 

r  =  r,  =  \(2n  -  5  ± 


+  4»  -  2)- 


(77) 


In  all  these  cases  the  panel  point  at  foremost  end,  having 
the  greatest  moment  as  the  load  advances,  lies  between  r^  and 
(r,  +  i)c,  except  when  rj  and  (r,  +  i);:  are  panel  points. 


42 


MECHANICS  OF   THE   GIRDER. 


s 

00        S-            °3-            £            %            g 

* 

-        ^            *            *             £            1 

* 

Jf             **                       ^t-                     VO                      CO                       O 

M                             N                             M                             CO 

* 

to           ON                   ON                 oo                   oo                   oo 

* 

O               O                        O                        O                        O               O 

O      N     vo 
M     r^.    ON 

"*  1  + 

« 

oo           -3-                oo                 M          vo           o 

M                                         M                                         tf                        "Si"                        -^i 

*JI_+ 

* 

VO               OO                        VO                rl-               N                          O 
i-O             VO                          to               d                CO                       VO 

N           Tf         Tj- 

CO      N        •tj- 

1   5. 

* 

M                          tO                       CO                                       N                                         N 

^  °?  ? 

* 

^>    §>       *        J        S 

ON                 ON 

hH                         HH 

i 

wwtoMto^           vovo            t^oo            ONO 
XXXXXX            XX            ><X            XX 

MhHMhHMCOONCOOMCOONtO 

. 

g 

666666666666666 

From  Equati 

H      tOVO      ^M       NVOOOOOOO      T)-TtO      O      O 

WP-H                        WW                       NNMCOCO 

XXX                  X                  X                  X 

j—  i      W               ro                      (*O                      ^5                      ro 

1    1    1    1    1    1    1    1    1    1    1    I    1    1    1 

IH                MM                COM                COM                tOM 

X            XX            XX            XX            XX 

ooooooooooooooo 

:aneous  moments 

•—  i 

SI8 

ooooooooooooooo 

1  N 

0) 

hHMMwNCOhiCO^MTj-l-OMiOVO 

O 

HU.. 

666666666666666 

»      % 

H 

'£        & 

CO      CO      Tt      rt      Tt 

I    % 

k 

M        M        N        M        W        fOM        CO^l-M        -<3-iOCO"1VO 

2       ^ 

* 

OOOOOO»HMMMNNCOfOco 

S        S 
'S       'x 

nj             rt 

1 

M      a      N      cocococococococococococo 

g           S 

MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   43 

We  now  give  an  example  of  an  equal-panelled  girder  trav- 
ersed by  an  odd  number  of  equal  apex  weights,  and  will  then 
illustrate  the  application  of  equations  (75),  (76),  and  (77). 

EXAMPLE.  —  Girder  of  10  panels  10  feet  each;  3  weights, 
of  8  tons  each,  at  intervals  of  10  feet  between  consecutive 
weights.  What  are  the  simultaneous  moments  at  all  the 
panel  points,  as  the  foremost  weight  of  this  load  passes  each 
panel  point  in  succession  ?  Use  equation  (63),  where,  now, 
W  =  8,  /  =  100,  n  =  10,  rz  =  o,  i,  2,  3,  etc.,  in  succession. 

Moments  within  the  same  brace,  simultaneous.  (See  table, 
p.  42.) 

Formula  gives  only  the  first  moment  in  any  horizontal  line 
of  moments.  For  other  moments  in  same  line,  add  four  times 
the  parenthetic  quantity  to  the  moment  immediately  before 
the  required  moment. 

For  the  greatest  moment  at  foremost  end  of  this  moving 
load,  we  have  from  (75),  where,  now,  e  =  3, 


rx  +  i  =  6, 

which  agrees  with  the  above  table  ;    the  moment  being   480 
when  the  foremost  end  of  load  passes  either  of  these  points. 
Also,  if  e  =  4,  as  in  example  4,  article  20,  we  have,  from  (75), 

r,  =  -  -  =  5.25  ;  and  5.50,  which  gives  the  greatest 

moment  576,  at  foremost  end  of  load,  lies  between  5.25  and 

6.25. 

For  a  full  load  coming  upon  the  panel  points  of  a  girder 
having  10  equal  panels,  (76)  gives  rt  =  |(io  —  i)  =  6,  r,  +  I 
=  7,  a  result  in  accord  with  the  solution  in  article  21,  where 
the  moment  at  foremost  end  is  greatest,  and  equals  672  at  these 
two  points. 


44  MECHANICS  OF   THE   GIRDER. 

Also,  when  n  =  10,  (77)  gives  rx  =  5.98,  rl  +  I  =  6.98  ;  and 
6.5,  giving  686  foot-tons  (example  of  article  23),  lies  between 
5.98  and  6.98. 

28.  To  find  the  general  expression  for  the  point  of  greatest 
moment,  on  the  loaded  part  of  the  beam,  due  to  a  uniform  dead 
load,  a  weight,  W,  being  applied  at  each  of  the  (n  —  i)  or  n 
panel  points,  and  to  a  uniform  live  load  consisting  of  rt  —  rz 
equal  weights,  Z,  applied  at  consecutive  panel  points  as  the 
load  advances,  we  employ  equation  (63),  putting  L  for  Wy  and 

rl 
x  —  —  =  re,  and  so  have  Mr.     Then,  substituting  r  -f-  i  for  r, 

we  get  Mr  +  „ 


for  the  loaded  part  of  the  beam. 

This  expression  added  to  (71),  and  the  sum  made  equal  to  o, 
gives 

L  \  n(n  -  i)^_  (n  -  ra)(r,  +  r2  +  i)  +  2«r,  }      (78) 

~T~      rr    }    (  ix 


to  be  used  when  the  value  of  r  lies  between  r2  and  rx  ;  and  the 
panel  point  under  the  live  load,  having  the  greatest  moment, 
lies  between  re  and  (r  -f-  i)^:  when  re  and  (r  +  i)c  are  not  panel 
points. 

In  a  similar  manner,  putting  r  =  r2,  and  .r  =  ?y:,  (r2  —  i)^, 
in  succession,  in  (63),  finding  &M^  and  adding  it  to-  AJ/in  (71), 
we  derive 

i  ~  ra)[2»  -(rf  +  r2  +  i)]         (79) 


where  r  is  not  greater  than  r2,  that  is,  at  the  left  of  a  partial 
live  load.     Also,  when  the  point  of  greatest  moment,  consider- 


MOMENTS   OF  FORCES  APPLIED    TO  BEAM  OR    GIRDER.   45 

ing  dead  and  live  loads,  lies  beyond  the  live  load,  we  derive, 
from  (63), 

,  -  r,)(rf  +  r2  +  i),  (80) 


where  r  is  not  less  than  r,,  that  is,  beyond  the  live  load. 

29.  We  next  assume  that  the  uniform  load,  «/  units  of  weight 
per  linear  unit  of  beam,  advances  by  continuous  increments, 
and  not  by  leaps,  or  entire  panel  weights  added  at  once ;  and 
require  the  moment  at  the  foremost  end  of  a  load  which  is  thus 
uniformly  distributed  continuously  from  its  foremost  end  to  the 
left  end  of  the  girder. 

Equation  (56)  applies  here  if  we  make  x  =  b  =  length  of 
uniform  load  measured  from  the  left  support ;  and  we  have 


(81) 


And,  if  w  is  the  unit  weight  of  the  dead  load,  we  have  from 
(49),  by  putting  x  =  b, 

Mbw  =  \wb(l-b},  (82) 

where  M^,  is  the  moment  of  a  beam,  at  the  distance  b  from 
one  extremity,  due  to  the  unit  weight,  w,  covering  the  entire 
beam. 

For  the  total  moment  due  to  live  and  dead  loads  at  the 
foremost  end  of  £ze/,  we  take  the  sum  of  (81)  and  (82),  and 
have 


(83) 


46 


MECHANICS  OF   THE   GIRDER. 


EXAMPLE.  —  Let  /  =  100,  w  =  0.4  ton,  «/  =  0.8  ton ;  and 
find  the  moments  at  the  foremost  end  of  the  moving  load,  bw'y 
at  intervals  of  10  feet  as  it  advances.  From  (83),  — 


b. 

\bl~b 

w'b  +  Iw. 

Mbii/  +  bw. 

t*  l  . 

o 

o 

40 

O 

10 

4.5 

48 

216 

20 

8.0 

56 

448 

3° 

10.5 

64 

672 

40 

12.0 

72 

864 

5° 

I2.5 

80 

IOOO 

60 

I2.O 

88 

1056 

70 

10.5 

96 

1008 

80 

8.0 

104 

832 

90 

4-5 

112 

5°4 

IOO 

0 

I2O 

o 

Each  of  these  moments  is,  as  it  should  be,  less  than  that 
found  for  apex  loads  by  just  the  moment  due  \w'  at  the  point 
taken,  since  the  point  at  the  end  of  the  continuously  distributed 
live  load  sustains  but  half  a  panel  weight  of  the  live  load. 

30.  If  it  be  required  to  find  the  moment  due  to  both  dead 
load,  Iw,  and  live  load,  bwf,  at  any  point  ahead  of  the  latter,  we 
use  for  the  live  load  (57)  by  making  a  =  o,  and  for  the  dead 
load  (49),  and  have 


Mx  =  - 

21 


(84) 


Or,  for  (r  +  i)  intervals,  each  equal  to  -,  we  find,  if  b  =  — , 

n  n 

(r  +  i)l 


+  i)  («  -  r  -  i).    (85) 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   47 


EXAMPLE.  —  Let  /  =  100,  it/  =  0.8,  w  =  0.4,  n  —  10 ;  and 
find  the  total  moment  at  the  distance  x  =  b  +  10,  or  at  the 

end  of  the  (r  +  i)th  interval,  -.  * 

n 

First  computation,  using  (84),  — 


b. 

jr. 

«/ 

2/' 

«•. 

l  —  X. 

First 
Term. 

%*>. 

l—x. 

X. 

Second 
Term. 

Mx. 

O 

10 

0.004 

0 

90 

O 

0.2 

90 

10 

180 

180 

10 

20 

0.004 

IOO 

80 

32 

0.2 

80 

20 

320 

352 

20 

30 

0.004 

400 

70 

112 

O.2 

70 

30 

420 

532 

30 

40 

0.004 

900 

60 

216 

O.2 

60 

40 

480 

696 

40 

50 

0.004 

1600 

50 

320 

0.2 

SO 

50 

500 

820 

50 

60 

0.004 

2500 

40 

400 

0.2 

40 

60 

480 

880 

60 

70 

0.004 

3600 

30 

432 

O.2 

30 

70 

420 

852 

70 

80 

0.004 

4900 

20 

392 

0.2 

20 

80 

320 

712 

80 

90 

0.004 

6400 

IO 

256 

0.2 

10 

9° 

1  80 

436 

90 

IOO 

0.004 

8100 

0 

0 

0.2 

O 

IOO 

O 

0 

Second  computation,  using  (85),  — 


r. 

trtl 

2«3' 

r2. 

«—  r—  i. 

First 
Term. 

w/2 
M* 

r  +  i. 

«—  r—  i. 

Second 
Term. 

^r  +  i. 

0 

4 

0 

9 

0 

20 

I 

9 

180 

180 

I 

4 

i 

8 

32 

2O 

2 

8 

320 

352 

2 

4 

4 

7 

112 

20 

3 

7 

420 

532 

3 

4 

9 

6 

216 

20 

4 

6 

480 

696 

4 

4 

16 

5 

320 

2O 

5 

5 

500 

820 

5 

4 

25 

4 

400 

2O 

6 

4 

480 

880 

6 

4 

36 

3 

432 

20 

7 

3 

420 

852 

7 

4 

49 

2 

392 

2O 

8 

2 

320 

712 

8 

4 

64 

I 

2S6 

20 

9 

I 

180 

436 

9 

4 

81 

0 

0 

20 

IO 

0 

o 

0 

This  second  computation  will,  in  general,  be  found  more 
convenient  than  the  first,  since  n  and  r  are  usually  integers, 
and  not  very  large. 


48  MECHANICS   OF   THE    GIRDER. 

31.  When  a  uniform  continuous  load  is  coming  upon  one 
end  of  a  girder,  we  may  find  the  position  of  the  foremost  end 
of  the  load  at  the  instant  the  moment  at  that  end  reaches  its 
maximum  value,  by  differentiating  (81)  with  respect  to  b,  and 

putting  —  —  -  —  o.     Thus, 
do 


do 

.'.     b  =  \L  (86) 

32.  The  position  of  the  foremost  end  of  the  live  load  when 
the  moment  is  a  maximum  there  for  combined  dead  and  live 

loads,  is  determined  by  differentiating  (83),  and  making  —  =  o. 

do 

Thus,  dM  =  »(2rfJ  _     ,.  +  /2  _  2/^  = 

do         2.1 

.'.     J  ^.Z,-  i  ±  (<«  +  <?  +  i)l,  (87) 


where  e  =  —  ,  and  b  =  length  of  live  load  on  one  end  of  the  girder. 
Equation  (87)  is  illustrated   by  the  example  in  article  29, 

where  e  =  —  ^-  =•  2  ;  and  (87)  gives  b  =  60.76,  while  (83)  gives 
0.4 

the  corresponding  moment  1056.31,  a  maximum. 

33.  Equation  (55)  expresses  the  moment  due  any  uniform 
partial  or  complete  continuous  load,  ixfb,  at  any  loaded  point,  x. 

By  differentiating  (55),  and  putting  -  =  o,  we  may  find  the 

dx 

value  of  x,  which  gives  the  maximum  moment.     Thus, 

-  20}  =  o, 


*  =  a  +  b  -b-(a  +  \b),  (88) 


which  is  the  point  of  greatest  moment  due  ix/b. 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   49 

34.  Also,  from  (49),  we  may  make 


and  find 

*  =  K  (89) 

which  is  the  point  of  greatest  moment  due  full  continuous  uni- 
form load,  as  in  equation  (52). 

35.  If  we  add  —  from  (55)  to  ^from  (49),  and  equate  the 
ax  dx 

sum  to  zero,  we  shall  find 

*  =  (a  +  *)wf  ~  (a  +  ¥}  ^    +        '    (90) 


which  is  the  point  of  greatest  moment  due  both  loads,. 
and  wl. 

36.  We  will  now  consider  the  case  of  a  girder  having  n 

panels,  each  =  c  =  -;  and  we  will  suppose  the  live  load  to 

consist  of  weights  not  all  equal,  nor  spaced  so  as  to  conform 
to  the  panel  points.  Such  a  case  is  presented  by  a  locomotive 
and  train  of  cars. 

»        Making  use  of  equations  (40)  and  (43),  let  us  arrange  for- 
mulae convenient  for  this  case. 

If  in  these  equations  we  put  nc  for  /,  and  for  a'  and  x  write 
the  proper  multiple  of  c,  we  have  the  simultaneous  moments 
due  each  weight,  W,  in  its  position  at  a  panel  point,  as  indi- 
cated in  the  following  tabular  arrangement  :  — 


50 


MECHANICS  OF   THE   GIRDER. 


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MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    51 

Now,  in  this  equation,  (91),  for  the  sum  of  the  moments  due 
all  the  weights,  we  may  evidently  put  any  weight  in  the  place 
of  any  other,  and  suppose  any  number  of  the  weights  equal  to 
zero. 

Hence  we  may,  by  means  of  (91),  find  the  momental  effects 
of  any  load  traversing  the  girder,  at  each  of  the  equal  intervals, 
c,  in  its  progress. 

EXAMPLE.  —  Let  the  span  =  I  =  100  feet ;  n  =  10  =  num- 
ber of  panels,   each  =  c  =  10  feet.      Dead  load  =  w  =  0.5 
ton  per  linear  foot  =  cw  =  5  tons  at  each  panel  point  or  apex, 
and  2j  tons  on  each  abutment.     Live  load  consists  of  two  loco-* 
motives,  each  of  the  following  lengths  and  weights  :  — 

Between  bearings  of  truck  wheels,  5.75  feet. 
Between  bearings  of  second  truck  wheels  and  first  driver,  8.50  feet  =  St. 

Between  bearings  of  drivers,  7.75  feet  =  S. 

Between  bearings  of  second  drivers  and  first  tender,  7.25  feet  =  S2. 

Between  bearings  of  first  and  second  tenders,  4.00  feet. 

Between  bearings  of  second  and  third  tenders,  7.25  feet. 

Between  bearings  of  third  and  fourth  tenders,  4.00  feet. 

Total  wheel  base,  44-5°  feet. 

Between  bearings  of  first  tender  and  truck  of  second  engine,  8  feet. 
Total  weight  of  tender,  42,000  pounds  =  21.00  tons. 

Total  weight  of  engine,  65,000  pounds  =  32.50  tons. 

Total  weight  on  2  pairs  drivers,  42,000  pounds  =  21.00  tons. 
Total  weight  on  2  pairs  truck,  23,000  pounds  =  11.50  tons. 
Weight  on  each  pair  truck  wheels,  5.75  tons  =  k. 

Weight  on  each  pair  drivers,  10.50  tons  =  D. 

Weight  on  each  pair  tender  wheels,  5.25  tons  =  /. 

Suppose  one  girder  carries  these  two  locomotives. 
We  first  find  the  greatest  weight  that  can  come  upon  a 
panel  point  or  joint  from  the  weights  in  adjacent  panels. 


52  MECHANICS  OF  THE   GIRDER. 

Let  Dl  =  D»  Fig.  10,  be  the  equal  weights  on  each  pair  of 
drivers,  and  take  A,  B,  C,  any  three  consecutive  joints  at  the 
given  interval,  c  feet ;  let  x  =  AE,  S  =  space  between  bases  of 
drivers,  it  being  less  than  c. 


x         V  ) 

s 

C) 

\                         E 

B 
FIG.  10. 

f 

d 

Then  the  weight  at  B,  from  drivers,  is 


X  ,-»  2C  —  X   —  S ,-»  2C  —  S  r^  /       \ 

-A  H A  =  Z>,  (92) 


which  is  a  constant,  while  the  point  B  is  anywhere  in  the 
space  5. 

If  both  drivers  are  between  two  consecutive  joints,  as  AB, 
we  have 

c  c  c 

which  is  not  a  constant,  but  reaches  its  greatest  value  within 
the  prescribed  limits  when  x  ==•  c  —  S ;  that  is,  when 


Therefore  — D  is  the  greatest  pressure  that  can  come 

upon  any  joint  from  the  drivers  of  one  engine. 

Now,  when  the  foremost  driver,  Dlt  is  at  B,  the  second  truck 
wheel  is  between  B  and  C;  and  when  the  second  driver  is  at 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    53 

B>  the  first  tender  wheel  is  between  A  and  B.     In  the  former 
case  the  increment  of  weight  at  B  from  the  truck  would  be 

-  l-k  ;  in  the  latter  case  the  increment  at  B  from  the  tender 
c 

c  —  S 
would  be  -  2-t.      Therefore  the  first  or  second  driver  at  B 

c  —  S, 
gives  the  greatest  pressure  at  that  point  according  as  —  -  —  k 

c-S2 
is  greater  or  less  than  —  -  —  /. 

In  the  present  example, 

c  —  Sl         10  —  8.5 

—f-k  =  —  —      x  5-75  =  0.86250, 

c  —  S2         10  —  7.25 

-—'    =    -       "    -      X      5.25     =     1.44375- 


We  will,  therefore,  find  the  pressures  at  points  whose  intervals 
are  equal  to  c,  when  the  second  driver  of  the  foremost  engine 
is  at  one  of  these  points.  Let  this  point  be  the  third,  counting 
from  the  right-hand  pier. 

Then  Fig.  1  1  shows  the  positions  of  all  the  wheels  with  ref- 
erence to  the  joints  at  the  equal  intervals  ;  and  a  simple  calcu- 

fc        fc  «     t  it  r(        p\  k        k 

B          Q     B.TJ    Q  8  Q    4  Q        7M        Q*O       7.H        Q      *>»       C)        ""SO        Q     ^8'Q  _  A 

I     3.75+5.75+.60     |          7.60  +  2.60         |     1.504  7.25*1i»   I  2.75+A26  \          7.«7B-«-2.Uft       |         8.  26  1  3.75  |  2^8  | 

a7  as  as  a4  a,  a2  a,  a 

t      t  t       t          Ji  X    n 

C  _  _  _  Q     4    Q        T.25        Q    4     O      7.2;     (    )        T.75        {)        B 

1  6+4.1-1  \         6-  26."*  3.75          |     .25*7.25  +  4.68     |        6.25  +  4.75  | 

a.,,  ato  a9  a.  a* 

FIG.  ii. 


lation  according  to  the  principle  involved  in  (38)  and  (39)  gives 
the  total  pressure  at  each  joint,  which  pressure  is  to  be  substi- 
tuted for  W  in  the  equation  (91). 


54  MECHANICS  OF  THE   GIRDER. 

In  this  position  of  the  locomotives  the  pressures  at  the 
equal  intervals  due  to  the  weights  on  the  adjacent  panels  are  — 

1.  At  A,  o.20o£  =     1.15000  tons. 

2.  Ata,,  1.425^  =    8.19375  tons. 

3.  At  a2,  0.375/6    +  0.775/7  =  10.29375  tons. 

4.  At  a3,  1.225/7  +  0.275*    =  14.30625  tons. 

5.  At  #4,  1.750*  =    9.18750  tons. 

6.  At  #s,  1.725*  =    9.05625  tons. 

7.  At  #6,  0.250*    -h  1.325^    =     8.93125  tons. 

8.  At07,  0.675^    +  o-525^  =    9-39375  tons- 

9.  At  08,  1.225/2  +  0.025*    =  I2-99375  tons* 

10.  At  09,   0.250/7+  i .600*    =  11.02500  tons. 

11.  At  0IO,  1.775*  =    9-3l875  tons- 

12.  At  «„,  0.600*  =    3.15000  tons. 

Total,  107.00000  tons. 

Stopping  with  the  second  driver  at  ay  we  see  that  the  hind- 
most tender  truck  has  not  yet  come  upon  the  girder.  We  com- 
pute, however,  the  moments  due  this  load  as  it  advances  panel 
by  panel,  till  the  twelfth  weight  is  upon  the  girder,  and  the 
first,  second,  and  third  have  passed  off;  using  indices  Mu 
Mt_2,  .  .  .  M3_II}  M4_12,  to  denote,  inclusively,  what  panel 
weights  produce  the  simultaneous  moments  opposite  M. 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    55 


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MECHANICS  OF  THE   GIRDER. 


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MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    57 


The  moments  for  dead  load,  given  opposite  Mw,  have  been 
computed  by  (65),  whilst  (91)  has  been  used  in  finding  the 
moments  due  the  advancing  load. 

The  differences  are  taken  directly  from  the  computed  mo- 
ments ;  and  we  must  evidently  use  for  each  half-span  the 
greatest  difference  due  any  interval,  the  load  being  supposed 
to  travel  either  way. 

37.  Let  us  now  suppose  that  this  same  live  load  of  107  tons 
is  distributed  uniformly  over  the  10  panels,  so  that  W  —  panel 
weight  =  10.7  tons.  We  then  find  by  means  of  (65)  the  great- 
est moments  due  live  load,  and  by  means  of  (69)  the  greatest 
differences  of  moment  due  live  load.  Taking  th,e  moments  due 
dead  load  as  found  above,  we  write  :  — 

WEIGHT  OF  Two  LOCOMOTIVES  UNIFORMLY  DISTRIBUTED. 


DISTANCE  FROM  PIER. 

10 

20 

30 

40 

50 

r. 

- 

I 

2 

3 

4 

^(r  +  i)r 

64  20 

2 

2« 

Full  live  load,  difference  .... 

481.50 
225.00 

374.50 

175.00 

267.50 
125.00 

160.50 
75.00 

53-50 
25.00 

4 
5 
6 

Maximum  difference  +     .     .    .     • 
Maximum  difference  —    .... 

706.50 
481.50 

549-50 

856.00 

392.50 
1123.50 

235-50 
1284.00 

78.50 
-82.00 
1337.50 

7 

Dead  load,  M     

225.00 

400.00 

525.00 

600.00 

625.00 

8 

Total  M  maximum  

706.50 

1256.00 

1648.50 

1884.00 

1962.50 

DISTANCE  FROM  PIER. 

60 

70 

80 

90 

100 

r. 

5 

6 

7 

8 

9 

Wc(r  +  ^)r 

224.7O 

385.20 

481.50 

2 

2« 

Full  live  load,  difference  .... 

—  53-5° 

—  I60.50 

-267.50 

-374-50 

—481.50 

4 
5 
6 

Maximum  difference  +     .     .     .     . 
Maximum  difference  —    .... 

-185.50 

-299.70 

—424.60 

856  oo 

—560.20 

48l.50 

-706.50 

Dead  load,  M     

225  oo 

3 

1884  oo 

164.8  c;o 

MECHANICS  OF  THE   GIRDER. 


A  comparison  of  these  maxima  moments  and  differences 
with  those  just  found  for  the  natural  distribution  of  the  weights 
of  these  two  locomotives,  shows  but  one  moment  and  one  differ- 
ence to  be  less  for  uniform  load  than  for  naturally  distributed 
load.  The  extreme  length  of  wheel  base  of  these  two  united 
locomotives  is  2  X  44.5  +  8  =  97  feet. 

It  will  be  observed,  that,  when  the  second  driver  of  the  sec- 
ond engine  is  at  a  joint,  the  weight  at  that  joint  is  greater  than 
the  weight  we  have  used  in  the  calculation  of  moments.  But, 
when  the  second  driver  of  the  second  engine  is  at  a  joint,  the 
second  driver  of  the  first  engine  is  2.5  feet  (see  Fig.  n)  from 
a  joint ;  so  that,  assuming  the  coupled  locomotives  to  travel 
either  way,  our  calculation  is  correct. 

We  will  close  this  section  with  an  example  including  every 
kind  of  loading  contemplated  herein. 

EXAMPLE.  —  Let  W  =.  20.0  tons,  a'  =    50  ft. 

w   =    0.4  tons,  /    =  100  ft. 

w'  =    0.8  tons,  b    =    20  ft.,  a  =  40  ft. 

P   =  10.0  tons,  a"  —    50  ft.,  a  =  30°. 

Find  the  moments  and  differences  of  moment  for  every  ten  feet 
throughout  the  girder. 

In  this  calculation  we  use  equations  (40),  (43),  (49),  (53), 
(55)»  (57)>  (47) >  and  (48)>  witn  tne  following  result :  — 


DISTANCE. 

10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

For  W,  M 

IOO 

200 

300 

400 

5°o 

400 

300 

200 

IOO 

0 

Forw,  M 

1  80 

320 

420 

480 

500 

480 

420 

320 

180 

o 

?OT'Z</,M 

80 

160 

240 

320 

360 

320 

240 

1  60 

80 

0 

ForP,  M 

-25 

-50 

—75 

—  IOO 

—  125 

—IOO 

—75 

-5°   . 

-25 

0 

Total     M 

335 

630 

885 

IIOO 

1235 

IIOO 

885 

630 

335 

o 

Total  dif.. 

335 

295 

255 

215 

135 

-135 

—215 

-255 

-295 

-335 

MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   59 


SECTION  3. 

Horizontal  Girder  of  One  Span,  with  Fixed  Ends.     Effects  of  End 

Moments. 

38.  If  we  suppose  the  simple  girder,  Fig.  9,  not  only  sup- 
ported at  its  ends,  but  also  fixed  by  being  built  into  the  walls, 
or  by  means  of  forces  applied  to  the  sections  AB  and  OE,  to 
keep  them  from  changing  place  as  the  beam  inclines  to  yield  to 
the  other  applied  pressures,  we  then  have  a  moment  developed 
at  each  extremity  of  the  girder,  which  will  manifestly  affect  the 
normal  moment  due  all  other  applied  pressures  at  every  cross- 
section. 

Let  us  now  find  expressions  for  the  momental  effects  at  any 
point  of  the  girder  due  to  the  given  end  moments,  without 
attempting  at  present  to  formulate  the  value  of  these  end  mo- 
ments, nor  to  determine  whether  they  are  simply  sufficient  to 
"fix"  the  ends  of  the  girder. 

Let  AB,  Fig.  12,  represent  a  beam  whose  end  moments  are 
M,  and  M2. 


M: 


FIG.  12. 


Call  the  length  of  clear  span  /,  and  take  VS  any  vertical 
section  at  the  horizontal  distance  ^  from  the  left  abutment. 
Let  Vj,  =  the  vertical  re-action,  positive  or  negative,  at  B,  due 


6O  MECHANICS  OF   THE   GIRDER. 

to  the  moment  Mt  at  A  ;   and  let  v2  =  the  vertical  re-action, 
positive  or  negative,  at  A,  due  to  the  moment  M2  at  B. 
Then,  taking  moments  about  A  and  B,  we  have 


Call  the  moment  at  VS, 

For  Vl,  -Vl(l  -  *)  =  --^M*  ; 
For  z/2,      zye  =       -J£. 

Hence  the  moment  at  FS  due  to  the  two  end  moments  acting 
in  opposite  directions  is 

Ttf  I  —    X  **•      ,,      .#  ji.f  JltFz    —    M-,  -»/r  f       \ 

Mx  =  ——Ml  +  -M2  =  -^—  -  -  l-x  +  Mlt      (93) 

where  both  end  moments  tend  to  diminish  the  normal  moment 
at  the  section  KS,  and  are  negative. 

If,  therefore,  we  apply  the  correction  (93)  to  the  moment 
found  at  any  cross-section  of  a  girder  with  free  ends,  which  we 
have  called  the  normal  moment,  we  shall  have  the  total  mo- 
ment, including  the  influence  of  the  end  or  pier  moments. 

39.  When  c  =  /  -f-  n  =  one  of  the  equal  panel  lengths  of 
the  girder  whose  end  moments  are  M^  and  MM  we  may  find  the 
momental  difference  for  any  interval,  c,  due  to  Ml  and  M2,  by 
putting  x  +  c  for  x  in  equation  (93),  and  subtracting.  Thus, 


c)  -f 


e  -  Mx  =  (J/2  -  ^)-        (94) 


By  means  of  (94)  we  may  correct  the  normal  difference  of 
moments  for  the  influence  of  the  given  pier  moments. 


MOMENTS   OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.   6 1 

The  pier  moments  Mt  and  M2  are  here  supposed  to  be  con- 
stant. The  cases  of  their  variation  will  be  considered  here- 
after, when  we  come  to  formulate  their  values. 

EXAMPLE  I.  —  Let  us  suppose  that  the  girder  for  which  we 
have  computed  the  maxima  moments  and  differences  of  mo- 
ment, in  article  25,  example  I,  had,  in  addition  to  the  pressures 
there  given,  been  subjected  to  these  end  moments ;  viz., 

Mt  =  —400  foot- tons, 
M2  =  —500  foot-tons. 

From  (93)  we  find  decrements  of  moment : 

400  —  500 

— x  —  400  =  —410  when  x  =     10 

IOO 

=  —420  when  x  =  20 

=  —430  when  x  =  30 

=  —440  when  x  =  40 

=  —450  when  x  =  50 

=  —460  when  x  =  60 

=  —470  when  x  =  70 

=  —480  when  x  =  80 

=  —490  when  x  =  90 

=  —500  when  x  =  100 

From  (94),  or  from  the  decrements  just  found,  we  have  the 
constant  decrement  of  difference, 

400  — 


Applying  these  corrections  to  the  tabulated  maxima  mo- 
ments and  differences  in  article  25,  example  i,  there  results  :  — 


62 


MECHANICS  OF  THE   GIRDER. 


X. 

0 

10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

M.    .    . 

—400 

130 

540 

830 

IOOO 

1050 

980 

790 

480 

50 

-500 

Difference     .    j 

53° 

418 

3H 

218 

+130 

+5° 

—7° 

-150 

-238 

-334 

-438 

-55° 

EXAMPLE  2.  —  Let  us  suppose  that  the  girder  of  example  in 
article  37  has  its  right  end  extended  20  feet  beyond  the  point 
of  support,  and  has  a  weight,  W,  =  10  tons  applied  at  that 
extremity.  What  is  the  pier  moment  developed  by  the  10  tons 
and  by  the  girder's  own  uniform  weight,  w  =  0.4  ton  per  linear 
foot  ?  And  what  is  the  effect  of  this  pier  moment  on  the 
normal  moments  and  differences  already  found  for  the  given 
pressures  ? 


From  (22),  moment  due  W7  is  —  Wl  =  —  10  X  20    = 

i                0.4  X  2O2 
From  (25),  moment  due  w  is  — ~wl2  = — 

Moment  at  right  pier  ==  M2  = 

Moment  at  left  pier  =  MI  = 

Whence,  by  (93),  we  have  corrections  of  moment, 


—200. 
-  80. 

—280. 
o. 


o  —  280 
100 


x  -j-  o  =  —   28  when  x  =     10 


56  when  x  = 
84  when  x  = 
112  when  x  = 
140  when  x  = 
1 68  when  x  — 
196  when  x  = 
224  when  x  = 
252  when  x  = 
280  when  x  = 


20 

3° 
40 

5° 
60 
70 
80 
90 
100 


MOMENTS  OF  FORCES  APPLIED    TO  BEAM  OR   GIRDER.    63 


From  (94),  correction  for  differences, 

(o  -  280)  x  TV%  =  -28. 


Applying  these  corrections  to  the  computed  normal  mo- 
ments and  differences,  we  find  :  — 


X. 

0 

10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

M  .  . 

0 

307 

574 

801 

988 

i°95 

932 

689 

406 

83 

-280 

Dif.  +  .  . 
Dif.-.  . 

3°7 

267 

227 

187 

107 

-163 

-243 

-283 

-323 

-363 

64  MECHANICS  OF  THE   GIRDER. 


CHAPTER    IV. 

STRAINS  IN  FRAMED  OR  BUILT  GIRDERS,  DEDUCED  FROM  THE 
MOMENTS  OF  THE  EXTERNAL  FORCES  AND  FROM  THE  SHEAR- 
ING-FORCES, AND  FROM  THESE  COMBINED. 

40.  BY  the  definition  of  statical  moment,  as  given  in  article  9, 
it  is  the  product  of  two  numbers,  —  one  representing  the  length 
of  a  straight  line,  the  other  the  amount  of  force  conceived  to  be 
applied  at  either  end  of  the  given  straight  line  or  lever  arm, 
and  to  act  in  a  line  at  right  angles  to  that  arm.  If,  therefore, 
H  is  the  force  or  strain,  and  h  the  lever  arm,  the  moment  is 


and  the  strain  (95) 


whose  line  of  action  is  perpendicular  to  the  arm  h. 

It  hence  appears  that  strains  are  deducible  from  moments  ; 
and,  in  order  to  apply  this  method  to  the  determination  of 
strains  in  girders  of  the  most  general  description,  let  Fig.  13 
represent  one  end  of  a  framed  girder,  consisting  of  triangular 
panels,  as  ABD,  CDF,  EFH,  etc.,  or  of  quadrilateral  panels, 
ABDC,  CDFE,  EFHG,  etc.,  whichever  we  choose  to  conceive 
them.  Let  the  horizontal  projection  of  each  panel  length,  AC, 
CE,  EG,  etc.,  BD,  £F,  FH,  etc.,  of  both  top  and  bottom  chords, 
be  equal  to  2c\  and  let  the  apices,  A,  C,  E>  G,  etc.,  be  horizon- 
tally projected  at  the  centres  of  the  horizontal  projections  of 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  65 

BD,  DF,  FH,  etc.,  respectively.  Let  the  inclinations  of  the 
segments  of  the  top  chord,  AC,  CE,  EG,  etc.,  to  the  horizon  be 
«„  «2,  «3,  etc. ;  those  of  the  segments  of  the  bottom  chord,  BD, 
DF,  FH,  etc.,  to  the  horizon  be  ft,  ft,  ft,  etc. ;  the  inclination 
to  the  horizon  of  a  Y  web  member,  as  BA,  DC,  FE,  etc.,  be  <£„ 
02>  $3>  etc.  ;  and  the  inclination  to  the  horizon  of  a  Z  web  mem- 
ber, as  AD,  CF,  EH,  etc.,  be  6a  02,  03,  etc.  :  each  angle  of  incli- 
nation of  chord,  a,  ft  to  be  measured  from  the  horizontal  drawn 
through  the  left  end  of  the  chord  segment,  and  each  angle,  <£, 
0,  to  be  measured  from  the  horizontal  through  the  lower  extrem- 
ity of  the  web  member,  as  in  trigonometrical  notation. 


FIG.  13. 

Assume,  further,  that  each  member  of  the  structure  is  capa- 
ble of  resisting  the  strain  that  may  come  upon  it,  either  of 
tension  or  compression ;  and,  for  distinction,  call  strains  in 
compression  positive,  and  tensile  strains  negative. 

Also,  the  simultaneous  forces  acting  at  each  apex  are  sup- 
posed to  be  in  equilibrium,  and  the  structure  at  rest.  All  the 
dimensions  of  the  skeleton  girder,  as  Fig.  13,  are  given,  and 
may  be  varied  so  as  to  represent  all  the  usual  forms  of  girder, 
as  illustrated  below. 

Let  P  symbolize  the  strain  along  any  segment  of  the  top 
chord  ;  U  the  strain  along  any  segment  of  the  bottom  chord  ;  Y 
the  strain  along  a  Y  web  member  whose  slope  is  </>,  as  defined 


66  MECHANICS  OF  THE   GIRDER. 

above,  and  length  v ;  and  Z  the  strain  along  a  Z  web  member 
whose  slope  is  0,  and  length  z.  Therefore  /  =  —  v  sin  (</>  —  0), 
and  q  —  -srsin  (180  —  (9  +  a)  =  ^sin  (B  —  a),  where  /  is  nega- 
tive, and  represents  the  line  drawn  from  any  upper  apex  per- 
pendicular to  the  chord  opposite  it,  and  q  is  positive,  and 
denotes  the  length  of  the  perpendicular  drawn  from  any  lower 
.apex  to  the  chord  opposite. 

Let  Hr  and  Hr+l  denote  the  two  simultaneous  horizontal 
strains  at  consecutive  apices,  whose  difference,  A//,  is  the  great- 
est of  all  differences  of  simultaneous  horizontal  strains  for  that 
interval ;  and  let  Mn  Mr+l,  be  the  corresponding  moments,  and 
7/r,  //r+I,  the  heights  or  vertical  distances  from  those  apices  to 
the  axis  of  each  chord  opposite.  Then  H  —  M  •—  h,  and  l±H 
is  the  horizontal  component  of  the  strain  developed  in  the  diag- 
onal or  web  member  for  the  interval  to  which  A//  belongs. 

41.  Suppose  that  the  greatest  moments  due  the  given  load- 
ing have  been  computed  for  the  vertical  section  at  right  angles 
to  the  plane  of  the  girder  through  each  point,  B,  A,  D,  C,  F, 
etc.,  Fig.  13  ;  that  is,  at  intervals  each  equal  to  c :  and  call  these 
moments  Mlt  M2,  My  etc.,  at  these  consecutive  intervals.  Also, 
if  the  simultaneous  moments  which  yield  the  greatest  difference 
of  horizontal  strains  at  any  two  consecutive  apices  are  different 
from  these  greatest  moments,  as  may  be  the  case  for  rolling 
loads,  suppose  such  moments  known. 

We  then  have,  from  the  figure  and  from  the  principles  of 
articles  10  and  3, — 


Strain  along  BD,  U^  =  Mt  -r  pl ;  along  AC,  Pl  =  M3 

DF,  U2  =  M4  -r  /2;  CE,  P2  =  Ms 

Fff,Us  =  M6  -f/3;  EG,  P3  =  M7 

Generally  U  -  Mr  -f  /;  P   =Mr  +  l    -g. 

Strain  along  AB,  Kx  =  ^ff  -f  cos  ^, ;  along  AD,  Zx  = 
CD,  Y2  =  ^ff  -T-  cos^2;  CF,  Z2  = 


Generally  Y   =  LrH  -f  cos  ^ ;  Z    -  Ar  +  x  H 


(96) 


cos02.       (97) 

cos  0.    \ 


STRAWS  IN  FRAMED   OR  BUILT  GIRDERS.  6/ 

42.  Shearing  Forces  and  Strains.  — If  through  any  mate- 
rial body  or  structure  we  conceive  a  plane  to  pass,  dividing  the 
body  into  any  two  parts  whatsoever,  and  assume  either  one  of 
the  two  parts  to  be  fixed  in  position,  while  the  other  part  slides 
or  tends  to  slide  in  any  direction  along  this  plane,  then  the 
force  acting  in  a  line  parallel  to  the  dividing-plane,  and  causing 
this  sliding  or  tendency  to  slide,  is  called  a  shearing-force,  and 
the  strain  on  the  particles  of  the  body  lying  in  this  plane, 
resisting  or  tending  to  resist  the  shearing-force,  is  called  the 
shearing-strain.  The  amount  of  shearing-strain  per  unit  of 
the  shearing-surface  is  its  intensity  ;  and  the  intensity  at  the  in- 
stant of  rupture  (that  is,  at  the  beginning  of  actual  sliding  of 
one  part  of  the  body  over  the  other  along  the  shearing-plane)  is 
the  breaking  shearing-strain,  and  is,  in  general,  peculiar  to  each 
kind  of  material,  and  must  be  determined  by  experiment. 

The  published  results  of  trustworthy  experiments  for  deter- 
mining the  ultimate  resistance  of  materials  to  shearing,  are  very 
meagre ;  and  the  following  table,  compiled  from  the  works  of 
two  of  the  best  authorities  I  know,  viz.,  Professor  Rankine  and 
Mr.  Bindon  B.  Stoney,  is  probably  as  worthy  of  confidence  as 
any  published  records  of  the  kind. 


68 


MECHANICS  OF   THE   GIRDER. 


TABLE    I. 

ULTIMATE  RESISTANCE  OF  MATERIALS  TO   SHEARING,  IN   POUNDS,  PER 

SQUARE  INCH. 


MATERIAL. 

Resistance  to 
Shearing. 

Remarks. 

Metals. 

Cast-iron      .     . 

27700  R. 

Tensile  strength  ranges  from  1  3400  to  29000.   R. 

Cast-iron      .    . 

(  17920  to  ) 
(  20160  S.  ) 

"  Substantially  its  tensile  strength."     S. 

Wrought-iron  . 

55059  s. 

Mean  of  5  tests  by  Mr.  Jones,  punching-plates. 

Wrought-iron  . 

50400  S. 

Mean  of  2  tests,  Mr.  Little,  hammered  scrap, 

inch  punch. 

Wrought-iron  . 

43456  s. 

Mean  of  4  tests,  Mr.  Little,  hammered  scrap, 

two-inch  punch. 

Wrought-iron  . 

50848  S. 

Bar,  0.5  x  3  inches,  punched  both  ways,  Mr. 

Little,  mean. 

Wrought-iron  . 

48160  S. 

2  bars,  1x3  inches,  punched  both  ways,  Mr. 

Little,  mean. 

Wrought-iron  . 

46144  S. 

Flanged  tire,  1.8  x  5  inches,  edgewise,  by  Mr. 

Little. 

Wrought-iron  . 

52192  S. 

Rivet,  I  inch,  Mr.  Clark.      Tensile  strength 

53760. 

Wrought-iron  . 

45696  S. 

Rivet,  I  inch,  2  plates,  Mr.  Clark. 

Wrought-iron  . 

49952  S. 

Rivet,  I  inch,  3  plates,  Mr.  Clark. 

Wrought-iron  . 

50000  R. 

Steel  .... 

63796  s. 

Kirkaldy,  rivet  steel,  tensile  strength  86450. 

Timber. 

Fir  .     .     . 

S 

I    d'                f             "R    1 

Fir,  red  pine    . 

(      500  to  ) 
1     800  R.  ) 

Fir,  spruce  .    . 

600  R. 

Fir,  larch     .    . 

(     970  to  ) 

• 

i    1700  R.  ) 

Oak     .... 

2300  R. 

Oak     .... 

4000  S. 

Across  grain,  Rankine's  deduction  from  Par- 

sons's  tests  of  English  oak  treenails. 

Ash  and  elm    . 

1400  R. 

Abbreviations :  R.,  Rankine ;  S.,  Stoney. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  69 

Instead  of  a  plane  cutting  the  body  into  two  parts,  we  may 
conceive  it  cut  into  two  separate  parts  by  any  cylindrical  sur- 
face, and  may  suppose  the  sliding,  or  tendency  to  slide,  to  be  in 
the  direction  of  the  generating  line  of  the  cylindrical  surface, 
as  in  the  case  of  a  cylindrical  punch. 

43.  From  the  definition  of  shearing-force,  it  follows,  that  if 
any  girder,  as  Fig.  13,  be  cut  by  a  vertical  plane  at  right  angles 
to  its  own  plane,  then  the  shearing  force  or  strain  at  this  ver- 
tical section  is  equal  to  the  algebraic  sum  of  the  vertical  com- 
ponents of  all  the  forces  impressed  upon  either  side  of  this 
vertical  plane.  And  these  two  algebraic  sums  of  the  vertical 
components  of  the  forces  impressed  upon  the  opposite  sides  of 
this  vertical  plane  will  have  contrary  signs,  and  be  numerically 
equal,  except  when  a  vertical  force  or  weight  is  applied  in  the 
vertical  plane  itself,  in  which  case  the  shearing-strains  on  oppo- 
site sides  of  the  shearing-plane  will  differ  by  the  value  of  this 
weight  applied  in  the  vertical  plane. 

Since  the  resultant  of  parallel  forces  is  simply  their  algebraic 
sum,  if  the  external  forces  applied  to  a  girder  are  all  vertical 
(that  _  is,  made  up  of  the  applied  weights  and  the  consequent 
vertical  resistances  of  the  piers),  the  shearing-force  on  either 
side  of  a  vertical  shearing-plane  is  merely  the  difference  be- 
tween the  sum  of  the  weights  and  the  re-action  of  the  pier  on 
that  side. 

If,  therefore,  S  denotes  the  shearing-force  on  either  side  of 
the  shearing-plane,  W  being  positive  and  denoting  any  weight, 
and  V  being  the  vertical  re-action  and  negative,  on  the  side 
chosen,  we  then  have 

S=V  +  ^0W,  (98) 

where  2*  W  is  the  sum  of  all  the  weights  between  the  shearing- 
plane  and  the  point  of  support  having  the  re-action  V\  that  is, 
of  all  the  weights  on  the  length  x. 


MECHANICS  OF  THE   GIRDER. 


In  case  of  the  semi-beam  for  the  free  end,  V  =  o,  and 


Sum  of  weights  on  /  —  x,  S  = 


J-X 


W. 


Sum  of  weights  on  /, 


S  = 


(99) 


^  being  measured  from  the  fixed  end. 

When  the  girder  is  supported  at  both  ends,  the  re-actions 
due  to  a  single  weight,  W,  applied  at  the  distance  af.  Fig.  9, 
from  the  left  support,  are,  by  equations  (38)  and  (39), 


At  left  support,     V^  —  —  W 


I -a' 


At  right  support,  V2  =  —  W— ; 


calling  them  negative. 

And  for  any  number  of  different  weights  applied  at  different 
points, 


(100) 


Therefore  for  this  case  the  shearing-strain  at  a  vertical  section 
distant  x  from  the  left  support  is 


or 


(101) 


If  upon  the  girder  supported  at  both  ends  there 


are  «  —  i 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


equal  weights,  W,  at  equal  intervals,  c  =  -,  and  J  W  upon  each 
end  of  the  girder  on  a  pier,  we  have 


I=  --W  =  ra, 

2 


=  -»w  +  (r  +  \W  =  \W(2r  -  n  +  i)  ; 


(102) 


the  shearing-plane  being  at  the  rth  point  of   division  counted 
from  the  left. 

For  a  uniform  continuous  load,  !w,  upon  a  girder  supported 
at  its  extremities, 

r,  =  -\iw  =  ra.  | 

At  any  point,  x,  (I03) 

*S  =  —\lw  -f-  wx.  \ 


For  a  uniform  continuous  load,  Iw,  upon  a  semi-girder  at 
any  point,  x,  measured  from  the  fixed  end,  the  shearing-strain 
is 

S  =  (/-*)«,;] 
and  when  x  =  o,  (104) 


For  any  partial  uniform  continuous  load,  bwr,  Fig.  9,  on  a 
beam  simply  supported  at  its  two  ends,  the  re-actions  of  the 
piers  are 


(105) 


which  re-actions  are  identical  with  the  shearing-strains  for  the 
unloaded  parts  of  the  beam. 


MECHANICS  OF   THE   GIRDER. 


But  for  the  loaded  part  b,  the  shearing-strain  is 

•V)  +  it/(x  —  a), 


or 


S  =  - 


(106) 


If  a  =  o,  and  x  =  £,  (106)  becomes 

(io7) 

24 

which  is  the  shearing-strain  at  the  foremost  end  of  a  uniform 
continuous  load  reaching  to  the  left  end  of  the  beam  supported 
at  both  ends.  And  equation  (107)  gives  the  greatest  positive 
and  the  greatest  negative  value  of  5  for  this  kind  of  load ; 
since,  in  equations  (106),  x  cannot  be  greater  than  b,  and  in  the 
first  of  those  equations  5  is  an  increasing  function  of  x,  while 
in  the  second  5  is  a  decreasing  function  of  x. 

The  shearing-strain  at  any  point,  x,  of  the  partial  uniform 
continuous  load  on  a  semi-beam,  Fig.  8;  is 

S  =  it/ (a  +  &  —  x);  (108) 

x  being  measured  from  the  fixed  end,  and  not  being  greater 
than  a  +  b,  nor  less  than  a. 

In  order  to  simplify  the  application  of  equations  (101)  for 
the  important  case  of  a  partial  or  complete  uniform  discontinu- 
ous moving-load,  Z,  to  be  applied  at  equal  intervals,  c  —  -, 

n 

along  the  girder,  we  proceed  as  in  article  20,  where  we  found 
the  moments  due  such  a  load  on  a  beam  supported  at  both 
ends.  Let  r,  —  r2  =  the  number  of  weights,  Z,  on  the  beam 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  73 

at  any  instant.  Take  r  not  less  than  r2  nor  greater  than  r,. 
Then,  in  the  first  of  equations  (101),  we  have 

2a'°  =  rl-r2 
and 

V  =  <-[(ra  +  i)  +  (ra  +  2)  +  (ra  +  3)  +  •  •  •  r,] 

=  ^.(r,  -  ra)(r,  +  r2  +  i) 

for  the  first  term.  But  in  the  second  term,  2*L,  we  must  take 
L  no  times  for  the  left  unloaded  end  of  the  beam,  r  —  r2  times 
for  the  loaded  part,  and  rt  —  rz  times  for  the  unloaded  part  on 
the  right  end.  Therefore 

S  =  -y[(rr  -  ra)/  -  f-(rx  -  r2)(r,  +  r2  +  i)]       (109) 
for  the  shearing-strain  left  of  the  load. 

S  =  -  j[(rf  -  ra)/  -  ^(r,  -  ra)(rf  +  r2  +  i) 

-(r-ra)/J,     (no) 
which  is  the  shearing-strain  between  the  points  re  and  (r  +  i)^. 

S  =  —(r,  -  ra)(rx  +  r2  +  i)  =  -^(r,  +  0       (m) 

2/  2W 

if  r2  =  o,  and  c  =  - ;  and  this  is  the  shearing-strain  at  and 

n 

beyond  the  foremost  end  of  a  uniform  discontinuous  load  reach- 
ing back  to  the  left  end  of  the  beam. 

44.  The  influence  of  end  moments  on  the  normal  shearing- 
strains  may  be  regarded  as  operating  upon  that  term  only  of 
the  shearing-strain  which  expresses  the  re-action  of  the  pier. 

Now,  the  pier  moment  MM  acting  at  the  right-hand  pier,  will 

M 

affect  the  re-action  V^  of  the  left  pier  by  the  amount  — -*  ;  and 

the  pier  moment  Mu  acting  at  the   left   pier,  will  affect  the 


74  MECHANICS  OF   THE   GIRDER. 

M 
reaction  V2  of  the  right  pier  by  the  amount  --  -1.     But,  by  the 

principles  of  article  10,  the  force  --  —  ',  acting  at  the  right  end 

JI/T 
of  the  lever  arm,  /,  induces  a  re-action,  +  -—  *,  at  the  left  end  of 

that  arm,  that  is,  in  this  case,  at  the  left  pier,  where,  conse- 
quently, 


Similarly 


(112) 


which  are  the  increments  of  the  shearing-strains  due  to  the 
end  moments,  and  are  to  be  added  algebraically  to  the  shear- 
ing-strains found  for  the  given  load  on  the  same  beam  simply 
supported  at  its  two  ends. 

The  values  of  M^  and  M2  are  here  arbitrary,  but  will  be 
determined  for  particular  cases  in  subsequent  chapters  of  this 
work. 

45.  To  find  the  Shearing-Strain  at  any  Vertical  Section 
of  a  Girder  (Fig.  13)  in  Terms  of  the  Vertical  Components 
of  the  Forces  which  are  impressed  upon  the  Shearing- 
Plane  through  the  Members  of  the  Girder  cut  by  that 
Plane.  —  Using  the  notation  already  given  in  article  40,  Fig.  13, 
and  equations  (3),  we  have,  as  the  vertical  component  resulting 
from  all  the  pressures  on  the  left  of  each  odd  vertical  plane, 
or  the  planes  through  the  lower  apices,  B,  D,  Fy  etc., 

Left  of  B,  S0  =     Vlt 

Left  of  D,  S2   =  —Pl  sin  a,  +  Zx  sin  0,  -  U,  sin  ft  ; 

Left  of  F9  S4   =  -P2  sin  «2  +  Z2  sin  02  -  U2  sin  ft  ; 

Left  of  (27-+  i)thapex,  S2r  =  -Prsmar  +  Zrsm0r  -  £/rsinft;  (113) 
counting  r  on  P,  Z,  and  U. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  75 

And  on  the  left  of  each  even  vertical  plane,  or  those  through 
the  upper  apices,  A,  C,  E,  etc., 

Left  of  A,  St  =  —P0  sin  «0  —  F,  sin  <£T  —  £/,  sin  /?T  ; 
Left  of  C,  S3  =  —  />  sintf,  -  F2sin<£2  -  £4  sin  ft  ; 
Left  of  £,  Ss  =  —  P2  sin  «2  —  F3  sin  <jf>3  —  £/3  sin  @3  ; 

Left  of  (2r)th  apex, 

1  —  Frsin<£r  —  £7rsin/?r;    (114) 


counting  r  on  F  and  U. 

These  values  of  5  may  be  used  to  verify  solutions  by  equa- 
tions (96)  and  (97),  as  will  be  illustrated  in  some  of  the  exam- 
ples below. 

46.  Strains  in  all  Members  of  a  Girder  determined  from 
the  Given  Shearing-Forces.  —  Equilibrium  of  the  system  re- 
quires that  at  each  apex,  Fig.  13,  the  sum  of  the  horizontal 
forces,  as  well  as  the  sum  of  the  vertical  forces,  shall  vanish. 
Therefore  at  any  lower  apex  we  have 

C7r_lcosftr_I  —  Zr_1cosOr_I  —  Yr  cos  <f>,  —  Urcos(3r  =  o,    (115) 
and 

!  +  Zrcos6r  +  Frcos<£r  —  Prcosar  =  6     (116) 


at  any  upper  apex. 

The  four  equations,  (113),  (114),  (115),  (u6),  enable  us  to 
determine  the  four  quantities,  Ur,  Pr,  Yr)  Zr,  in  terms  of  />r_J, 
and  the  given  shearing-strains,  S2r_^  and  .S2r,  if  we  use  the 
auxiliary  equation, 

—  Pr-i  cosar_l  =  [7r_Icosfir_I  —  Zr_IcosQr_1,     (117) 

expressing  the  equality  of  the  horizontal  strains  at  any  lower 
apex,  and  at  the  point  directly  above  it  in  the  top  chord. 


76  MECHANICS  OF   THE   GIRDER. 

Therefore,  after  solving  and  reducing, 

—  Pr  _  i  sin  (<J)r  —  ar  _ 


I -y,  sin  (ft. -ft.)  ,  ) 

/  f\                    \  ^  *' 

sm(0r  —  «r) 

•_r   —  £^  sin  ft  —  ^.x  /  x 


•*7"  •          j  /  \  / 

sm$r 

7     —  ^  Sin  «r  +  ffi-  sin  ft  +  «S*2r  (l2l) 

sin0r 

Now,  if  we  begin  at  the  left  end  of  the  girder,  Fig.  13,  to  com- 
pute, Ur  becomes  U»  and  Pr_t  is  zero;  therefore  (118)  gives 
Ul :  and  with  this  value  of  Ur  =  Ul  we  at  once  find  Pr  =  PJt 
by  (119),  and  similarly  follow  Yr  and  Zr  from  (120)  and  (121). 
A  repetition  of  this  process,  putting  the  value  of  Pr  just  found, 
in  the  place  of  Pr-i>  may  be  continued  through  the  girder. 

47.  We  will  now  give  examples  illustrating  the  determina- 
tion of  strains  in  open  girders,  first  by  the  method  of  moments, 
and  second  by  the  method  of  shearing-strains,  and  will  verify 
the  solutions  by  equations  (113)  and  (114). 

EXAMPLE  i.  —  Let  B,  Fig.  13,  represent  the  unsupported  end 
of  a  semi-girder,  whose  fixed  end  coincides  with  the  vertical  plane 
passing  through  E,  and  at  right  angles  to  the  plane  of  the  girder. 
Let  the  horizontal  distance  between  consecutive  apices,  B,  A, 
D,  C,  etc.,  =  c  —  10  feet,  and  the  elevation  of  the  apices,  in  feet, 
above  the  point  B  be  as  shown  in  the  first  line  of  the  solution 
below.  These  elevations,  with  the  horizontal  distance  c=io feet, 
furnish  all  the  angles  and  lines  required.  If  any  apex  is  below 
B,  its  elevation  is  negative ;  and  all  angles  and  trigonometric 
functions  follow  the  ordinary  trigonometrical  laws.  At  each 
apex,  top  and  bottom,  of  this  semi-beam,  let  a  weight,  W—  i  ton, 
be  applied.  Required  the  strains  due  this  load  in  every  member 
of  the  girder.  The  moment  M  is  given  by  (35),  where  ;/  =  5, 
^  —  5°>  ^=  !»  and  r  takes  the  values  4,  3,  2,  i,  o,  in  succession. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  *JJ 


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MECHANICS  OF  THE   GIRDER. 


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STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


79 


LOGARITHMIC  SOLUTION  BY  EQUATIONS  (118)  TO  (121). 
Method  of  Shearing-Strains. 


r. 

1 

2 

3 

<br 

60°  q6'  4V 

62°  14'  ^o" 

6l°  76'  2  C" 

o 

•;°42/4i;/ 

2°  U'4V 

60°  c6'  4V 

S6°  ii  '40" 

c8°  44'  40/x 

0r    . 

2°  51'  45" 

4°  if  21" 

o 

<br         Br 

q8°    4'  c8" 

c7o  r7'    Q" 

6l°T.6'2C" 

180°  —  6r    ~ 

1  80°        (Or        Br) 

59032'   4" 

62°2V4Q// 

6o°i5/i8// 
64°  -j''  ^o" 

l8o°        (Or        Or)      

6=;°  14'  4V 

6V>   7'   V 

i 

5 

log^W-i      

log  cos  <br                              •     • 

0. 

9  6863166 

0.477I2I3 
9.6681466 

0.6989700 
9  67  7  1  666 

log  Szr  -  1  cos  <br 

9.6863166 

O  I4CJ267Q 

o  7761766 

Szr  -  1  cos  <i>r     •     .          ... 

0.48  1;64 

1  .  7Q7  2 

2.7776 

log/V-i       .     .          

o.  2  240067  n 

o.74526i9« 

Q.Q2i2t;8t; 

Q  07180^8 

log/V-isin(<pr  —  ar-x)  .... 
—  /V-isin  (<£r  —  OT-I)      .     .     .     . 

0.48564 

0.1452652^ 
1-3972 
2.7944 

0.677  1  577» 
4-7551 

7.1727 

lognum  

log  sin  (<t>r       0r)        •          «     «          « 

9.6863166 

Q  Q288l  IQ 

0.4462948 

Q.Q28  IQ1^ 

0.8532540 

Q  0447777 

log£/r      

0.7  C7  CQ47 

O.^lSoQQ1^ 

0.008016"? 

Ur 

O  t\72I 

•3.2060 

8  1081 

S2r  

2 

4 

log^r      

O.3OIO3OO 

0.6020600 

0.70^02  <2n 

96oc;6oi8« 

log  Szr  cos  Or     •     •     •          • 

0.00605  52W 

o  2976618;* 

S*r  cos  Or      

—  1.01404 

—  1.084  q 

log  sin  (Or       @r)    •          • 

Q  Q4.7  C2I4. 

Q  Q  C  C6478 

log  Ur  sin  (Or       Pr)  

Q  7OW26I 

0.4777477 

—  Ur  sin(0r  —  ftr)    

—  o.  ^0702 

—  2.0768 

—  1.52106 

4.Q6l  7 

lognum  

0.1821464;* 

0.69  5  w  c  qw 

O.Qc8l'JQ7 

O.QCO7776 

log/V      

o  2  240067  n 

O  74  ^26lQW 

Pr  

1,674.0 

c  c6^C 

log  sin  OT  -i  

8.QQ78007 

8.6984422 

8o 


MECHANICS  OF   THE   GIRDER. 


LOGARITHMIC  SOLUTION  BY  EQUATIONS  (118)  TO  (121). —  Concluded. 


r. 

1 

2 

3 

9.22i9o64« 

0.4477041  n 

—0.16666 

O  27777 

8.6984422 

8.8778446 

log  UT  sin  ftr    

8.4.CCQ46Q 

Q.7QIQ44I 

o  0080167 

Ur  sin  /3r      

0.028^7 

0.246^7 

o 

—  1.028^7 

—  7.O7QO 

4  72^2 

log  num  . 

0.012277972 

o  488538072 

Q.Q4I  "$Q  I 

Q  Q46QO4O 

Q  Q447777 

logFr.     . 

0.0706448;* 

o  ^41674072 

o  7  298087  w 

Kr  

I  1766 

•7  480? 

From  Pr  -  1  sin  ar  -  1  comes  Pr  sin  or, 
Numerator  of  (121)  .     .     .     . 

-0.16666 
I  8619 

J-4UU0 
-0.27777 
7  0688 

O'jWy 

o  2600^87 

j.yuoo 

o  59861581 

log  sin  0r      

Q  Q7  ^4741 

Q  Q7864O8 

logZr      

O  77A/4.8/l6 

Zr  

2  1601 

4o/  A1 

N-B-  —  The  method  of  shearing-strains,  though  applicable,  is  not  conveniently 
used  for  live  loads,  since  every  change  of  load  requires  recomputation  from  the 
beginning. 

48.  Maxima  Strains  in  the  Web  Members  of  an  Open 
Girder,  deduced  from  the  Moments  and  Shearing-Forces 
combined,  for  Uniform  Discontinuous  Dead  and  Live 
Loads. 

Let  W    —  panel  weight  of  dead  load  at  (n  —  i)  equidistant 

points, 
L      =  panel  weight  of  live  load  to  be  applied  at  the 

same  points, 
Mw  =  moment  at  each  panel  point  due  dead  load,  by 

equation  (65), 
ML  =  moment  due  live  load  at  its  foremost  end,  by 

equation  (64), 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  8  1 

Sw  —  shearing-force   at   each   panel   point   due   dead 

load,  by  (102), 
SL    =  shearing-force  at  foremost  end  of  live  load  due 

live  load,  by  (in)  ; 

.*.     Sw  +  SL  =  greatest  shearing-force  simultaneous  with  MW  -f-  ML, 

—  w     —  -  =  H  =  simultaneous  horizontal  chord  strain. 
h 

Now,  we  have  on  the  immediate  right  of  any  vertical  plane- 
through  an  upper  apex  (Fig.  1  3), 

C/cosfi  =  —Hr  =  -Pcosa  -f  Zcos0, 
—  Psina  +  ZsmO  —  £7  sin  ft  =  Sr. 

Whence,  after  eliminating  U  and  P,  there  results, 


cos/3 


where  H  and  5  belong  to  the  vertical  section  through  the  upper 
extremity  of  the  ZQ  member,  which  joins  the  left  end  of  the  Pa 
chord  segment,  to  the  right  end  of  the  Up  chord  segment. 

Similarly,  on   the   immediate   right   of  the  vertical   plane 
through  the  consecutive  lower  apex, 


cos  a 


•  .,  cos 


where  fft  and  5X  belong  to  the  given  vertical  plane  through 
the  lower  extremity  of  the  Y  member,  which  joins  the  left 


82  MECHANICS  OF  THE   GIRDER.  . 

end  of  the  Ulft  chord  segment,  to  the  right  end  of  the  Pa  chord 
segment  of  equation  (122). 

It  may  here  be  observed,  that  according  to  our  notation 
'(article  40,  Fig.  13),  in  any  symmetrical  girder,  0  in  either  half- 
span  is  the  supplement  of  <£  in  the  corresponding  panel  of  the 
other  half-span ;  also  a  and  ft  in  either  half-span  are  respec- 
tively equal  to  — a  and  —  ft  of  the  corresponding  panel  of  the 
other  half-span.  So  z  of  the  first  half-span  equals  the  corre- 
sponding v  of  the  second. 

EXAMPLE.  —  Uniform  discontinuous  dead  and  live  loads. 
Let  all  that  part  of  Fig.  13  which  is  on  the  left  of  the  vertical 
line  through  E  represent  one  of  the  equal  half-spans  of  a  girder 
supported  at  its  two  ends,  B,  and  L  not  shown  in  the  figure. 
Take  the  dimensions  for  each  half-span  the  same  as  those 
already  given  in  the  example  of  article  47. 

Let  the  dead  load,  W^  =  4  tons,  be  applied  at  each  apex,  top 
and  bottom ;  and  the  live  load,  Lt  =  8  tons,  at  the  same  points 
progressively.  Each  extreme  apex  may  be  supposed  to  bear 
J(  W  +  L)  when  fully  loaded ;  but  this  will  only  affect  the 
resistances  Vv  and  V2,  so  far  as  the  present  method  of  com- 
puting strains  is  concerned.  We  may  find  greatest  strains  as 
.follows :  — 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


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84 


MECHANICS  OF  THE  GIRDER. 


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86 


MECHANICS  OF  THE   GIRDER. 


The  chord  strains,  U  and  P,  are  to  be  found  as  before  ;  their 
values  being  greatest  when  the  two  uniform  loads  cover  the 
beam. 

In  the  second  line  of  this  last  solution,  ML  is  the  moment 
due  live  load  at  its  foremost  end  as  that  end  passes  the  suc- 
cessive apices. 

In  the  third  line,  M+^L  is  the  moment  one  interval  beyond 
the  foremost  end,  and  simultaneous  with  ML. 

It  is  manifest,  from  what  precedes,  that  we  need  compute 
the  moments  MWt  ML,  and  M+lL,  only  when  h  is  not  constant ; 
as,  when  h  does  not  vary,  we  may  find  &MW  and  &ML  by  (71) 
and  (69),  whence  A//  =  &M  -H  h. 

49.  We  now  proceed  to  classify  girders  according  to  the 
form  which  the  general  equations  assume  when  particular 
values  are  assigned  to  one  or  more  of  their  variables ;  first, 
recapitulating  the  general  equations  of  the  method  of  moments, 
and  of  the  method  of  moments  and  shearing-forces. 

From  equations  (95),  (96),  (97),  (122),  (123),  we  arrange 

GENERAL  FORMULAE. 


METHOD  OF 

Moments. 

Moments  and  Shearing-Forces. 

P       = 

n(,-0  =  -*rcoB0. 

p 

=  —v  sin(0  —  ft)  =  —  hr  cos/3. 

9       = 

z  sin(0  —  a)  =  hr+i  cos  a. 

q 

=  z  sin(0  —  a)  =  hr  +i  cos  a. 

H    = 

±M 

±h. 

H 

=  ±(Mw+Mj-)  +A. 

&H  = 

Hr+i. 

—  Hr,  or 

S 

—  <S  if?  -f-  «S^« 

&H- 

Mr  +i       Mr 

P 

M                                       H 

tir+i 

hr 

M(W+L)(r+i-)  .  ^     //(W+^)(r+i)  •  cc 

p     — 

Mr  + 

c  _  Hr+\ 

u 

—  yj/                '  i)  —  H                '  cos  8 

q 

cos  a 

(W+L)r  .P           (W+L)r  . 

TT     - 

Mr 

-Hr 

Hr  +i  sin(/3t  —  a)  —  Sr+i  cos  a  cos  /3X 

P 

'  cos/3* 

cos  a  sin(0  —  /3X) 

Y    = 

brH 

-fcos^. 

—  Hr  sin  (/3  —  a)  +  Sr  cos  a  cos  ft 

Ar+i 

H  -i-  cos  0. 

cos/3  sin(0  —  a) 

STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  8/ 

v          =  length  of   the    Y  web  member  making  the  angle  <£ 

with  the  horizon,  Fig.  13. 
z  =  length  of  the  Z  web  member  making  the  angle  0  with 

the  horizon. 
p          =  length  of  perpendicular  drawn  from  any  upper  vertex 

to  the  lower  chord. 
q          =  length  of  perpendicular  drawn  from  any  lower  vertex 

to  the  upper  chord. 

h          —  height  of  girder  at  any  apex. 
a          =  inclination  of  any  segment  of  the  upper  chord  to  the 

horizon,  as  angle  CAM. 
(3          =  inclination   of  any   segment  of   the  lower  chord  to 

the  horizon,  as  angle  FDN. 
<£          =  inclination  to  horizon  of  any  Y  web  member,  as  angle 

CDN. 
0  =  inclination  to  horizon  of  any  Z  web  member,  as  angle 

ADN. 

W        =  panel  weight  of  dead  load. 
L          •=.  panel  weight  of  live  load. 
MW      —  moment  due  dead  load. 
ML       =  moment  due  live  load  at  its  foremost  end. 
MW+L  =  moment  due  dead  load  and  full  live  load ;   that   is, 

greatest  moment  for  uniform  loads. 

H         =  horizontal  component  of  chord  strain  at  a  joint  or  apex. 
&H      =  difference  of  simultaneous  horizontal  components  of 

chord  strains  at  consecutive  apices  when  this  dif- 
ference is  greatest. 
Sw      =  shearing-force  due  dead  load  on  the  immediate  right 

of  the  shearing-plane. 
SL        =  shearing-force  due  live  load  at  any  point  beyond  its 

foremost  end. 

P         —  strain  in  any  segment  of  top  chord. 
U         =  strain  in  any  segment  of  bottom  chord. 
Y         =  strain  in  any  Kweb  member. 
Z         =  strain  in  any  Z  web  member. 

Count  r  always  from  the  left,  as  indicated  in  the  figures. 


88 


MECHANICS  OF  THE   GIRDER. 


Now,  although  we  have  thus  far  considered  each  upper  ver- 
tex to  be  horizontally  projected  midway  between  the  horizontal 
projections  of  the  lower  vertices,  this  restriction  is  by  no  means 
necessary  in  the  application  of  these  equations,  provided  we 
compute  the  moments  and  the  shearing-strains  in  accordance 
with  the  distribution  of  the  loads,  whatever  that  may  be. 

THE  TWELVE  CLASSES  OF  GIRDERS  OF  SINGLE  SYSTEM. 


Length. 

a. 

V. 

z. 

c. 

Strain. 

P. 

Y. 

Z. 

U. 

Tension 

Compression 

Class. 

Top  Chord. 

a. 

Web 

*. 

Web 

9. 

Bottom  Chord. 

0. 

Member. 

Member. 

I.      .     . 

Inclined     . 

a 

Inclined  . 

<A 

Inclined  . 

0 

Inclined 

3 

II.    .     . 

Inclined     . 

a 

Inclined  . 

</> 

Inclined  . 

d 

Horizontal, 

o 

III..     . 

Horizontal, 

0 

Inclined  . 

0 

Inclined  . 

e 

Inclined 

0 

IV.  .    . 

Horizontal, 

0 

Inclined  . 

0 

Inclined  . 

d 

Horizontal, 

0 

V.    .    . 

Inclined     . 

a 

Inclined  . 

<t> 

Vertical  . 

90° 

Inclined     . 

(8 

VI.  .    . 

Inclined     . 

a 

Vertical  . 

90° 

Inclined  . 

e 

Inclined     . 

f 

VII.     . 

Inclined     . 

a 

Inclined  . 

* 

Vertical  . 

90° 

Horizontal, 

o 

VIII.    . 

Horizontal, 

0 

Inclined  . 

f 

Vertical  . 

90° 

Inclined 

a 

IX..    . 

Horizontal, 

0 

Inclined  . 

$ 

Vertical  . 

90° 

Horizontal, 

0 

X.    .    . 

Inclined     . 

a 

Vertical  . 

90° 

Inclined  . 

e 

Horizontal, 

0 

XL  .    . 

Horizontal, 

0 

Vertical  . 

90° 

Inclined  . 

e 

Inclined     . 

0 

XII.     . 

Horizontal, 

o 

Vertical  . 

90° 

Inclined  . 

e 

Horizontal, 

o 

The  conditions  yielding  the  twelve  classes  may  be  briefly 
stated  thus :  — 

neither,      1 


With  regard  to  a  and  /3  we  may  have 


the  one, 


the  other,      =  °;    4  conditions. 


both, 

(neither,      1 
the  one,      I  =  90°  •    3  conditions, 
the  other,  J 

Combining  these  conditions  gives  twelve  classes  and   no 
more. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


89 


CLASS  I.  —  ALL  MEMBERS  BUT  ONE  INCLINED. 

Use  General  Formulae. 
P3 


FIG.  14.  —  THE  CRESCENT  GIRDER. 
P2 


FIG.  15.  —  THE  TRUNCATED  CRESCENT. 

Pa 


FIG.  16.  —  THE  DOUBLE  Bow,  OR  BRUNEL  GIRDER. 


FIG.  17.  —  INVERTED  TRUNCATED  CRESCENT. 


9o 


MECHANICS  OF  THE   GIRDER. 


FIG.  18.  —  INVERTED  OR  SUSPENDED  CRESCENT. 


FIG.  19.  —  ROOF  PRINCIPAL. 


FIG.  20.  — ROOF  PRINCIPAL. 


FIG.  2i.  — BENT  GIRDER  OF  FOUR  SYSTEMS. 


STRAIN'S  IN  FRAMED   OR  BUILT  GIRDERS.  9 1 


FIG.  210.  —  FIRST  SYSTEM. 


FIG.  2ib.  —  SECOND  SYSTEM. 


FIG.  2ic.  —  THIRD  SYSTEM. 


FIG.  zid.  —  FOURTH  SYSTEM. 


92 


MECHANICS  OF  THE   GIRDER. 


FIG.  22. —  DOUBLE  Bow  OF  Two  SYSTEMS. 


FIG.  220.  —  FIRST  SYSTEM.    FIG.  16  SHOWS  THE  SECOND  SYSTEM. 


FIG.  23.  — SEGMENT  OF  ROOF  PRINCIPAL.    (SEE  FIG.  71.) 


FIG.  230.  —  DIAGONALS  IN  COMPRESSION. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


93 


FIG.  23^. —  DIAGONALS  IN  TENSION. 


FIG.  24.  —  PARALLEL  CHORDS.    TRIANGULAR. 


FIG.  25.  —  THE  BRACED  ARCH.     ST.  Louis  BRIDGE  SYSTEM. 
(SEE  ARTICLE.) 


94 


MECHANICS  OF   THE   GIRDER. 


V 

Although  we  have  supposed  the  linear  dimensions  of  the 
girder  known,  we  will  now  give  a  mode  of  finding  them  from 
the  known  length  /,  and  central  height  //,  in  case  of  girders 
having  either  chord,  or  both  chords,  circular  or  parabolic. 


FIG.  26. 


ist,  Lower  chord  horizontal,  and  upper  chord  circular,  as 
ABCED,  Fig.  26.  Let  /  =  2DC  —  length  of  bottom  chord, 
h  =  AD  =.  central  height  of  girder.  Then  the  radius 


(124) 


Take  D,  the  centre  of  the  chord  of  any  arc,  as  ABC  or 
AJ$£t  for  the  origin  of  rectangular  co-ordinates  ;  the  axis  of  x 
being  horizontal,  and  that  of  y  being  vertical.  Then  the  equa- 
tion to  the  curve  ABC  is 


and 


(y  +  R  -  hY  =  R2  -  x2, 
=  y  =  h-  R  ± 


x)(R  -  x),        (125) 


which  is  the  height  of  the  bowstring  girder  at  any  point,  E, 
whose  distance  DE  from  the  origin  is  x. 


STRAINS  IN  FRAMED    OR  BUILT  GIRDERS.  95 

Thus,  if  /  =  ioo,  and  h  =  10,  we  have  R  =  IO°2  +  4  X  io2 
—  130;  and  from  (125)  we  find, — 

When  x  =    o,  y  =  10  =  h. 

x  =  10,  y  =  9.6148. 

x  =  20,  y  =  8.4523. 

x  =  3°>  J  =  6.4912. 

x  =  40,  ^  =  3-6931- 

*  =  45>  J  =  I-963I- 

x  =  5°  =  V>  y  =  °- 

The  length  of  any  diagonal,  DBy  whose  inclination  to  the 
horizon  =  <£,  is 

BE        DE 

V  =  -: — 7  =  1 ' 

sin  <j>       cos  <^ 

or,  in  general, 

v  =  yr  -r-  sin<^  =  &x  •+•  cos<£,  (126) 

&c  being  one  panel  length. 

The  length  of  any  diagonal,  AE,  whose  inclination  to  the 
horizon  =  0,  not  acute,  is 


z  = =      DE  . 

sin  0  cos  0 ' 

or,  in  general, 

.-jV-.  +  stotf--^.  (I27) 

The  length  of  any  chord,  AB,  whose  inclination  to  the  hori- 
zon =  a,  is 

a=  DE  =  ^  • 

cos  a       sin  a ' 

or,  in  general, 

a  =  -^-  =  -^L.  (128) 

cos  «       sin  a 


96  MECHANICS  OF  THE   GIRDER. 

From  (126), 

From  (127), 

^L  =  ->_,  +  A*  =  tan  0.  (130) 

cos  0 

From  (128), 

HL^L  —      AJF    +  A*  =  tana.  (131) 

cos  a 

Whence  </>,  0,  and  a,  and  their  sines  and  cosines,  may  be  taken 
from  a  table  of  natural  or  logarithmic  circular  functions. 

To  determine  the  length  of  any  part  of  the  circular  arc 
ABC,  or  of  the  whole  arc,  find  the  angle  at  the  centre  corre- 
sponding to  the  given  chord  of  the  required  arc ;  then  the 
required  length  of  arc  is  to  the  whole  circumference  as  the 
angle  at  the  centre  is  to  four  right  angles.  Thus,  if  C  denotes 
the  angle  at  the  centre  whose  chord  is  the  span  /  =  100,  we 
have 


.-.     C    =  45°  14'  23"  =  450.239722. 

Circumference  =  360  degrees. 

Length  of  circumference  =  2-n-R  =  2  X   3.14159  X    130 
=  816.8134, 


/.     Required  arc  2ABC  =      <  x  816.8134  =  102.645. 

Or,  the  length  of  the  arc  subtended  by  a  given  chord  may  be 
taken  from  a  table  constructed  for  the  purpose. 

2d,  Both  chords  or  flanges  circular,  as  ABCB.A,,  Fig.  26, 
where  the  chords  meet  at  the  ends  of  the  girder  ;  or,  as  Figs. 
15  and  17,  where  the  cjiords  do  not  meet  at  the  ends. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  97 

If  the  curves  meet,  as  in  Figs.  14,  16,  18,  and  26,  then  /  in 
equation  (124)  will  be  common  to  both  arcs,  ABC,  A1B1C\  but 
the  central  heights  of  the  two  arcs  will  be  h  =  AD,  and 
hi  =  A^D.  Hence,  for  the  upper  curve,  the  radius 


If,  as  in  Fig.  15,  the  curves  do  not  intersect  at  the  ends,, 
then,  for  each  curve,  /  will  be  the  chord  subtended  by  the  arc, 
h  will  be  the  central  height  of  each  arc  above  its  chord,  and  the 
origin  of  co-ordinates  for  each  curve  will  be  at  the  centre  of  its 
own  chord. 

The  ordinates  y,  corresponding  to  the  same  values  of  x,  are 
to  be  found  for  each  curve  by  (125)  ;  and  if  yt  =  an  ordinate  to 
the  upper  curve,  and  y  ==  the  corresponding  ordinate  to  the 
lower  curve,  and  e  =  the  difference  in  height  of  the  two  ori- 
gins, then  j,  -f-  e  —  y  =  the  height  of  girder  at  any  point,  x. 
And  when  e  =  o,  as  in  Fig.  26,  the  height  of  the  girder 
ABCB.A,  at  any  point,  B,  is  BB,  —  7,  —  y. 

FB  =  BE  -  AD  =  Ay,  in  general. 
F,B,  =  B,E  -  A,D  =  &ys,  in  general. 
DE  =  AF  =  A,Ft  =  A#,  in  general. 

Ay  A_y, 

tan«  =  -,     tanai  =  ~ 

AA,  +  BF  =  ylr     -  yr+I, 


y,r      -  yr 
tan(9  =  --  -  -  ,    tan  A  =   -~  -- 


98  MECHANICS  OF  THE   GIRDER. 

From  these  tangents,  a,  «„  0,  <£,  and  their  sines  and  cosines, 
are  to  be  found  as  before.     We  then  have 


yl        —  yr  &x 

•  -  4*  -  "S*—  =    c^' 


.    =  A,B    =  ^^p  =  -—ti,  (133) 

sm  6  cos  & 

a   =AB    =  £L  _-£*-,  (134) 

sm  a  cos  « 

x  T-»  Ay,  A^  /       x 

a,  —  A^Bi  =   -r^-  =      .  (135) 

cosaj 


3d,  When  the  curvature  of  one  or  both  chords  of  the  girder 
is  parabolic,  we  proceed  as  in  case  of  the  circular  chords  just 
discussed,  except  in  finding  the  ordinates  and  length  of  the 
curve,  which  only,  therefore,  we  need  now  determine. 

Let  the  curves,  Fig.  26,  now  be  parabolas,  whose  vertices 
are  at  A  and  Al  respectively.  Take  the  origin  of  rectangular 
co-ordinates,  as  before,  at  D  ;  the  axis  of  x  being  horizontal,  and 
that  of  y  vertical.  Then  the  equation  to  the  curve  ABC  is 

y   =(i-4|!^;  (I36) 

to  the  curve  A^B.C, 

(I37) 


For  the  same  value  of  x,  (136)  and  (137)  give 

-i)i  (138) 


which  is  the  height  of  the  girder  at  any  point  whose  distance 
is  x  from  the  centre  or  origin. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  99 

Thus,  if  /  =  ioo,  and  h  =  10,  (136)  gives,  — 

When  x  =    o,  y  =  10  =  h. 

x  =  10,  y  =     9.6. 

x  =  20,  y  =     8.4. 

x  =  30,  y  =     6.4. 

x  =  40,  ^  =     3.6. 

*  =  45>  ^  =     i-9- 

x  =  5°  =  K  y  =   o- 

And  if  /*,  =  ^,Z>  =  20,  A  =  ^Z>  =  10,  and  /  =  2DC  =  ioo, 
equation  (138)  gives  the  heights  of  girder  ACAt  as  below :  — 

When  x  =    o,  h0    =  10  =  /tt  —  h\ 

x  =  10,  hm  =  9.6  ; 

x  =  20,  ^20  =  8.4  ; 

x  =  3°>  ^30  =  6.4  j 

x  =  40,  /;40  =  3.6  ; 

X  =    45.  '     ^45   =       I-9> 

^  =  50  =  -|/,  7zso  =    o ; 

which  are  the  same  as  the  heights   of  the  girder  ADC  just 
found,  since  /it  =  2/1,  and,  from  (136)  and  (137), 


that  is,  the  ordinates  to  the  two  curves,  for  the  same  value  of  x, 
are  proportional  to  their  central  heights. 

The  length  of   the  parabolic  arc,  in  terms  of  the  chord  / 
and  the  central  height  //,  is 


(140) 


where  log  means  the  common  logarithm. 


IOO  MECHANICS  OF   THE    GIRDER. 

If  /  =  loo  feet  =  span,  and  h  —  10  feet  =  central  height, 
of  parabolic  arc,  then  (140)  gives  5  =  102.606  feet. 

From  these  examples  it  appears,  that,  when  the  curvature  is 
small,  there  is  but  little  difference  between  the  ordinates  and 
arc  of  the  circular  and  the  ordinates  and  arc  of  the  parabolic 
girder  of  the  same  central  height  and  span. 

Instead  of  these  exact  determinations  of  the  linear  dimen- 
sions of  a  girder,  the  figure  may  be  drawn  to  a  scale,  and  the 
length  of  each  member  measured,  where  greater  accuracy  is  not 
required. 

It  is  proper  to  observe  here,  that,  in  all  cases  of  curved 
flange,  the  line  of  action  of  the  flange  strain,  P  or  U,  is  the 
chord  of  the  arc  between  adjacent  apices,  and  not  the  arc  itself. 
When,  therefore,  either  flange  of  a  girder  is  curved,  and  not 
polygonal,  there  is  developed  midway  between  adjacent  apices 
in  the  same  flange  a  deflecting  force  tending  to  increase  the 
curvature  of  a  compressed  flange,  and  to  dimmish  the  curvature 
of  a  flange  in  tension. 

For  the  amount  of  this  deflecting  force  F  we  have,  if  P  is 
the  strain  along  the  chord  BIC,  Fig.  26, 

F  =  2P\a&HCL  (141) 

Or,  if  C  is  the  angle  at  the  centre  of  the  circle  whose  chord  is 
BC,  then 

i  -  cos 


/.     F  =  2/>(cosec£C  -  cot|C)  ;  (142) 

and  the  strain  along  the  chord  HC  of  each  half  of  the  arc  BC 
is 

P'=  P  H-  cosZfCY.  (143) 


STRAfJVS  IN  FRAMED   OR  BUILT  GIRDERS.  IOI 

Similarly,  in  cases  like  Pv  Fig.  19,  there  is  a  deflecting  force 
generated  at  the  ridge  equal  to 

F  =  2/>3  tan  a; 

and  the  strain  along  the  upper  segment  of  each  rafter  is,  as 
indicated,  P3  -r-  cos  a. 

The  bending-moment  due  to  this  deflecting  force  is  given 
by  equation  (46), 

M  =  ±Fa,  (144) 

where  a  is  the  length  of  the  chord  BC. 

The  amount  of  material  required  to  neutralize  this  moment 
will  be  determined  in  the  sequel.  But  it  is  already  manifest 
that  the  least  amount  of  resisting  material  will  be  required 
when  the  line  of  pressure  coincides  with  the  axis  of  the  resist- 
ing member. 

Multiple  or  compound  web  systems,  as  those  represented  in 
Figs.  21  and  22,  may  be  separated  into  the  single  systems  of 
which  they  are  composed,  when  the  sum  of  all  the  strains  found 
for  the  same  member  will  be  the  strain  sought  for  that  member. 


CLASS  II.  —  BOTTOM  CHORD  HORIZONTAL,  OTHER  MEMBERS 
INCLINED.      3  =  o. 


U,  u2  u3  u4 

FIG.  27.  —  THE  PARABOLIC  OR  CIRCULAR  BOWSTRING. 


102 


MECHANICS  OF   THE   GIRDER. 


U2  U3 

FIG.  28. —  THE  TRUNCATED  BOWSTRING. 


U2  U3 

FIG.  29.  —  ROOF  PRINCIPAL. 


U2  U3 

FIG.  30.  —  ROOF  PRINCIPAL. 


FIG.  31.  —  TIED  RAFTERS. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  103 


U2 
FIG.  32.  —  ROOF  PRINCIPAL. 


FIG.  33.  —  ROOF  PRINCIPAL  OF  Two  SYSTEMS. 


FIG.  ^a.—  FIRST  SYSTEM. 


FIG.     &  —  SECOND  SYSTEM. 


104 


MECHANICS  OF  THE   GIRDER. 


u, 

FIG.  34.  —  ROOF  MAIN,  COMPOUND  SYSTEM.     (SEE  FIG.  70,  CLASS  VIII.) 


FIG.  35.  —  BOWSTRING  OF  Two  SYSTEMS.     TRIANGULAR. 


U2,  U3  U4  U6 

FIG.  350.  —  FIRST  SYSTEM.    FIG.  27  SHOWS  THE  SECOND  SYSTEM. 


FIG.  36.  —  THE  POST  TRUSS  WITH  CURVED  TOP. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  105 


As  Counters.  As  Mains. 

FIG.  360.  —  FIRST  SYSTEM. 


1  2  3 

As  Counters.  As  Mains. 

FIG.  36^.  —  SECOND  SYSTEM. 


FIG.  37.  —  DOUBLE  TRIANGULAR  TRUNCATED  BOWSTRING.    SYSTEM  OF 
KANSAS-CITY  BRIDGE. 


FIG.  370.  —  FIRST  SYSTEM. 


io6 


MECHANICS  OF   THE   GIRDER. 


U2  U3 

FIG.  yjb.  —  SECOND  SYSTEM. 


FORMULA  FOR   CLASS   II.     /3  =  o. 

Method  of  Moments. 
H  =  M  -r-  h. 


XT' 


cosoc 
U  =  -^,. 


cos 


z_  Ar  +  1^ 


In  case  of  vertical  end  posts,  as  in  Fig.  37^,  where  Fand  Z 
become  indeterminate  by  the  above  equations,  we  have 


x  =  Z^  sin  0I  -f  Pl  sin  «„ 


=  F5  sin 


P4  sin  «4. 


Where  a  vertical  section  through  any  apex  cuts  a  web 
member,  as  in  Fig.  33$  for  Zl  =  F3,  and  in  Figs.  36^  and  36^ 
for  P4  and  the  counters  F  and  Z,  we  do  not  have  H,  the  hori- 
zontal component  of  chord  strain,  equal  to  M  -h  //,  but  may 
proceed  as  follows  :  — 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


ID/ 


Find  the  moments  Miy  M2,  M3,  etc.,  at  vertical  planes 
through  consecutive  apices.  Call  the  heights  above  the  bottom 
.chord  at  which  the  cut  diagonal  meets  these  vertical  planes  in 
each  panel,  a  and  b>  as  in  Fig.  38. 


FIG.  38. 

Then,  taking  moments  about  A  and  B,  we  have 
Mj.  —  h^  cos  a,  +  a,  Yl  cos  <£„ 
M2  =  /i2Pl  cos  «!  -h  bl  Yl  cos  </>„ 
p  _ 

•    •  -L\      


(146) 


(148) 


(aji2  —  £(AI)COS£J 

Taking  moments  about  A^  and  B»  there  results 

M2  =  —  h^ £/,  cos  ft  —  (/^2  —  ^)  ^i  cos(/>,,       (I5°) 
...     ^/;  —  — ii-i ^—  — ^-.  (151) 


I08  MECHANICS  OF   THE   GIRDER. 

Similarly,  or  by  increasing  the  indices  of  a,  b,  a,  and  0  by 
i,  and  those  of  M  and  h  by  2,  we  find 


(aji±-  bji^  cos  p2 
Then,  taking  moments  about  C  gives 


(152) 


Now,  in  case  of  the  Post  Truss,  we  need  only  the  counter- 
strains  F,  since  Z,  P,  and  U  have  their  greatest  values  as  main 
strains. 

And  when  both  chords  are  horizontal,  the  horizontal  pro- 
jection of  the  Z  member,  or  strut,  is  one-third  of  the  horizontal 
projection  of  the  Y  member,  or  tie,  as  usually  built  ;  hence 
a  =  \b  =  \h, 


*•  -          -  Sf 

A  TT 

which,  it  will  be  seen,  is  the  same  thing  as    Fx  =  -  ,  pro- 

cos  </> 

vided  A//  is  taken  for  the  interval  equal  to  the  horizontal 
projection  of  the  F  member,  while  ^H  belongs  to  one-third  of 
that  interval  ;  since  the  foremost  end  of  the  live  load,  at  the 
instant  the  value  of  F  is  here  sought,  is  at  the  foot  of  the  F 
member,  being  applied  either  directly  at  the  lower  apex,  or 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


109 


indirectly  at  the  upper  apex,  and  reaching  the  bottom  chord 
through  the  Z  member  terminating  there.  In  other  words, 
when  the  counter-strain  F,  due  live  load,  is  greatest,  there  is 
no  part  of  the  live  load  applied  on  the  right  of  the  foot  of  the 
Y member,  or  of  the  top  of  the  ^member  above. 

In  multiple  systems,  where  the  chords  are  not  straight  lines, 
in  finding  total  chord  strains,  care  should  be  taken  to  reduce  all 
strains  that  are  to  be  added,  so  that  their  lines  of  action  will 
be  parallel ;  horizontal,  for  instance. 

CLASS  III. — TOP  CHORD  HORIZONTAL,  OTHER  MEMBERS 
INCLINED,     a  —  o. 


FIG.  39.  —  TRUSSED  BEAM. 


FIG.  40.  —  TRUSSED  BEAM. 


P, 


Pi 


FIG.  41.  —  INVERTED  BOWSTRING. 


no 


MECHANICS  OF  THE   GIRDER. 


FIG.  42.  —  INVERTED  TRUNCATED  BOWSTRING. 
FORMULA  FOR   CLASS   III.     a  =.  o. 

Method  of  Moments. 
H  =  M  -h-  h. 


P  =  Hr 
U  =  — 


r+  cos  /. 
f-  cos  <j>. 
f  ^-  cos 


CLASS  IV.  —  BOTH  CHORDS  HORIZONTAL,  WEB  MEMBERS 
INCLINED,     a  —  o,  (3  =  o. 

Pi  P2  P3 


u,  U2  u3  U4 

FIG.  43. —  THE  TRIANGULAR  GIRDER.  » ERECT,  OR  "THROUGH." 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  Ill 


U2  U3 

FIG.  44.  —  TRIANGULAR  GIRDER.    SUSPENDED,  OR  "DECK."     a  —  o,  /?  =  o. 


FIG.  45.  —  DOUBLE  TRIANGULAR  GIRDER.    FIGS.  43  AND  44  COMBINED. 


FIG.  46.  —  QUADRUPLE  TRIANGULAR  SYSTEM.    EACH  SYSTEM  INDEPENDENT. 


U3 


FiG.'4&*.  —  FIRST  SYSTEM. 


112 


MECHANICS  OF   THE   GIRDER. 


FIG.  46^.  —  SECOND  SYSTEM. 


U, 


FIG.  46^.  —  TiiRD  SYSTEM. 


FIG.  46d.  —  FOURTH  SYSTEM. 


FIG.  47.  — THE  POST  TRUSS.    Two  SYSTEMS. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  113 


U7 


Z7 


As  Counters.  As  Mains. 

FIG.  470. —  FIRST  SYSTEM.    • 


U,  U2  U 


U4  U5  U6 

As  Counters.  As  Mains. 

FIG.  47^.  —  SECOND  SYSTEM. 


FIG.  47<r.  —  POST  TRUSS.    THREE  SYSTEMS. 
FORMULA  FOR  CLASS   IV.     a  ~  o,      =  o. 

Method  of  Moments. 
H  =  M  -*-  h. 

=  bM  +  h.  ((  UNIVERSITY 

•n  TJ 

f   =  Jjp  +  i« 

U  =  -Hr. 

Y  =  krH  -f-  cos  <^>. 
Z  =  Ar  +  iH  -r-  cos  0. 


114 


MECHANICS  OF   THE   GIRDER. 


CLASS  V.  —  ALTERNATE  WEB  MEMBERS  VERTICAL.     BOTH 
CHORDS  INCLINED. 

Generally  {   Verticals  in  compression,   \    Q  =      „ 
(   Diagonals  in  tension.    -      j 

Only  one  set  of  diagonals  shown  in  figures.    These  are  counters  in  first  half-span, 
mains  in  second  half-span. 


Ui  U2 


Us  U4  U5  Ua  U7  U8 

FIG.  48.  — THE  CRESCENT. 


FIG.  49.  —  TRUNCATED  CRESCENT. 


FIG.  50.  —  DOUBLE  Bow,  OR  BRUNEL  GIRDER. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  115 


FIG.  51.  —  TRUNCATED  CRESCENT  INVERTED. 


U4  U6 

FIG.  52.  — CRESCENT  SUSPENDED. 


VJi 


FIG.  53.  —  ROOF  PRINCIPAL. 


FIG.  54.  —  TRUSSED  RIB. 


MECHANICS  OF  THE   GIRDER. 


FIG,  55.  —  DOME  PRINCIPAL.    PRIMARY  SYSTEM.    SECONDARY  SYSTEM 
SAME  AS  IN  FIG.  54. 


Z  = 


FIG.  56.  —  BENT  TRUSS.    DOUBLE  SYSTEM. 


FORMULA  FOR   CLASS  V.     0  =  90°. 

Method  of  Moments. 
H    =  M  +  h. 
MfJ. 


-Mr. 

hr  +  T         hr 
P     =  H  -r-  cos  a. 

TT  TT  /D 

T»  ^      j-f 

—  Pr  sin  ar  —  Yr  sin  <j>r 

=  Hr  +  !  tan  ar  +  I  —  Hr  tan  ar  — 
(Load  applied  at  bottom.) 


—  Hr  tan  &•  +  //r  +  1  tan  /Jr  +  x  —  ArZTtan  <£r. 
(Load  applied  at  top.) 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


117 


The  value  of  Z  in  equation 

Z  =  kH  H-  cos  0, 


here  becomes  indeterminate,  since  &H  =  o  for  the  horizontal 
projection  of  the  Z  member,  and  cos  90°  —  o. 

CLASS  VI.  —  BOTH   CHORDS   INCLINED.     ALTERNATE  WEB 
MEMBERS  VERTICAL. 


»      (   Verticals  in  tension,  ) 

Generally   J  .  \   $  = 

(    Diagonals  in  compression.    ) 


90' 


Only  one  set  of  diagonals  shown  in  figures.     These  are  counters  in  first  half-span, 
mains  in  second  half-span. 


FIG.  57.  —  THE  CRESCENT. 


FIG.  58.  —TRUNCATED  CRESCENT. 


U4  U5 

FIG.  59.  —  THE  BRUNEL  GIRDER. 


MECHANICS  OF   THE   GIRDER. 


FIG.  60.  —  TRUNCATED  CRESCENT  SUSPENDED. 


FIG.  61.  —  SUSPENDED  CRESCENT. 


Y  =  Pr  +  1  sin  ar 


FORMULA  FOR   CLASS  VI.     </>  —  90°. 

Method  of  Moments. 
H     =  M  -j-  h. 

^r+I   _  Mr^ 

hr+i  hr 
P  —  H  -r-  cos  a. 
#_,  =  -H  +  cos  (3. 

—  Pr  sin  ar  —  Zr  +  x  sin  0r  +  1 
=  Z^.  +  !  tan  «r  +  !  —  ZTr  tan  ar  —  ^r 
(Load  applied  at  bottom.) 


ff  tan  O 


r  +  z. 


(Load  applied  at  top.) 
Z  =  &rH  +  COS  ^. 

Multiple  systems  of  this  class  are  seldom  built,  since  long 
struts  are  not  economical. 


STRAIArS  IN  FRAMED   OR  BUILT  GIRDERS.  1 19 

CLASS  VII.  —  BOTTOM  CHORD  HORIZONTAL.    fi  =  o.    ALTERNATE 
WEB  MEMBERS  VERTICAL.     0  —  90°. 


In  general,  j   Verticals  * 
(   Diagonals  i 


compression, 
in  tension. 


But  one  set  of  diagonals  shown  in  figures.    These  are  counters  in  first  half-span, 
mains  in  second  half-span. 


FIG.  62.  —  THE  BOWSTRING. 


U2  U3  U,  U5  Uff  U7  U8  U9 

FIG.  63.  —  TRUNCATED  BOWSTRING. 


FIG.  64.  —  RAFTERS  AND  TIE. 


I2O 


MECHANICS  OF   THE   GIRDER. 


~OT~  U3  U4  U5  U6  U,  U8  U9 

FIG.  65.  —  TOP  CHORD  UNIFORMLY  SLOPED.    ax  —  az  =  a3. 


U,  U2  U3  U4  U5  U9 

FIG.  66.  —  POLYGONAL  TOP  CHORD. 


U7  U8 


U2  U3  U4  U5  U6  U7 

FIG.  67.  —  ROOF  PRINCIPAL.    a^  —  a2  —  a3. 


U2        U3         U4  U5  Ue  U7  U8  Ufl        UIO       Utl 

FIG.  68.  —  PARALLEL  BRACING. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  121 


FIG.  69.  —  UNIFORM  SLOPE.     DOUBLE  SYSTEM. 


FIG.  70.  —  CURVED  TOP.    DOUBLE  SYSTEM. 


FIG.  71.  —  ROOF  PRINCIPAL.    (SEE  FIG.  23.) 


122 


MECHANICS  OF   THE   GIRDER. 


simultaneous. 


FIG.  jia.  —  THE  TWIN  FISHES,  LONG  SPAN.    (SEE  FIG.  22.) 
FORMULAE  FOR   CLASS  VII.     ft  =  o,  0  =  90°. 

Method  of  Moments* 
H     =  M  -f-  h. 

±H  =  Mr  +  *  -  ^ 
hr  +  i         hr 

P      =  Hr  -r-  cos  a. 
U      =  -Zfr  +  I. 

y    = 
z     = 

Foremost  end  of  live  load  at  Zr— i  for  maximum  Y  and  Zr> 

When  it  is  desired  to  have  the  diagonals  in  each  half-span 
parallel  for  a  given  number  of  panels,  as  in  Fig.  68,  the  lengths 
of  the  panels  and  the  inclination  of  the  diagonals  may  be  found 
as  follows :  — 

Let  /    =  length  of  span. 
7z0  =  height  at  each  end. 
h  =  central  height. 

m  =  number  of  panels  in  each  half-span. 
<£  =  inclination  to  horizon  of  counters  in  first  half-span, 

and  of  mains  in  second  half-span. 
a  =  inclination  of  top  chord. 
A/=  the  variable  panel  length. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


Then 


=  — — ,      and      AA  =  <_. 
tan  <£  tan  </> 


,  =  h  -  A*  =  h  - 


tan  </>          \         tan  <£ 


=  Aol  14-  — —  ,  , 
tand)/ 


tan<£  = 

Generally, 

hr  =  Ac 


=  h     ( i  —  tana>\  =  // 1  _  tan  « V 

" z  \         tan  <£/          \         tan  </>/  ' 

Ao    =  ^  _  ssjy, 


tan  a  /       x 

.-.     tan</>  = r  (JSS) 

'-(if 

for  the  first  half-span. 

Similarly,  for  the  second  half-span 


i  *  +  MA'.  (I58) 

\        tan  ^>/ 


124 


MECHANICS  OF   THE   GIRDER. 


CLASS  VIII.  —  TOP  CHORD  HORIZONTAL,     a  =  o.     STRUTS 

VERTICAL.       0  =  90°. 


FIG.  72.  — THE  FINK  TRUSS. 


U1 


FIG.  720.  —  MAIN  SUSPENDERS. 

P  P 


U 


u. 


FIG.  "j2b.  —  SECONDARIES. 


TERTIARIES. 


FIG.  73.  —  THE  BOLLMAN  TRUSS. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  125 


FIG.  74.  —  TRUSSED  RAFTER  OF  FIG.  34. 


FIG.  75.  —  SPANDREL  FILLED. 


U2 
FIG.  76.  —  TRUSSED  BEAM,  THREE  LINKS. 


R  P2 


P*  PS 


Pa 


P7  P8 


FIG.  77.  —  CATENARIAN  LINKS. 


126 


MECHANICS  OF  THE   GIRDER. 


FIG.  78.  —  TRUSSED  BEAM. 


FIG.  79.  —  THE  POINT  SUSPENSION,  STIFFENED  CATENARY. 


FORMULAE  FOR  CLASS  VIII.     a  =  o,  6  =  90°. 

4 

Method  of  Moments. 
H     =  M  -T-  h. 

hr  +  i         hr 


-Hr   -f-  COS  f3. 

-  cos</>. 


P 
U 

Y 
Z 


When  the  vertical  member  has  no  diagonal  attached  at  its 
top,  then,  of  course,  the  strain  upon  the  vertical  is,  for  Class 
VIIL,  equal  to  the  load  applied  at  the  upper  apex. 


P, 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  I2/ 


CLASS  IX.  —  BOTH  CHORDS  HORIZONTAL,     a  =  o. 

Verticals  in  compression.     f$  =  o. 
Diagonals  in  tension.     0  =  90°. 

P2  P3  P4  P5  P8  Pr 


U,  U2 


u3 


"07 


U8 


FIG.  80.  —  THE  PRATT  TRUSS. 


FIG.  Sew.  —  THE  LINVILLE,  OR  PRATT  OF  Two  SYSTEMS. 


FIG.  8o£.  —  PRATT  TRUSS  OF  THREE  SYSTEMS. 


128 


MECHANICS  OF  THE   GIRDER. 


For  end  struts,  tan  0  =  2  tan  <£. 
FlG.  SOC.  —  LlNVILLE,   WITH   INCLINED    END-POSTS. 


FIG.  Sod.  —  THREE  SYSTEMS,  INCLINED  END  POSTS. 


/ 
4 


FIG.  8o<?.  —  TRUSS  SYSTEMS  OF  NIAGARA  BRIDGE. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS. 


129 


U2  U3  U4  U6  U6  U7 

FIG.  8o/.  —  PRATT  TRUSS  SUSPENDED. 


FORMULA    FOR  CLASS  IX.     a  =  o,  ft  =  o,  0  =  90°. 


Method  of 


Moments. 


Moments  and  Shearing-Forces. 


H     =  M  -v-  h. 


P 

U       = 

Y      = 


H 


r-  cos  <£. 


H  = 


U  =  — 

Y   =  —S  -f-  sin</). 

Z   =  S+  x,  load  applied  at  top. 

Z  =  S,  load  applied  at  bottom. 


130 


MECHANICS  OF   THE   GIRDER. 


CLASS  X.  —  BOTTOM  CHORD  HORIZONTAL.    j3  —  o. 

Verticals  in  tension.     </>  =  90°. 
Diagonals  in  compression. 


U2  U3  U4  U5  U6  U7  U8 

FIG.  81.  —  THE  POLYGONAL  BOWSTRING. 


U,  U2  U3  U4  U5  Ue  U7  U8 

FIG.  82.  —  THE  HOWE  TRUSS,  WITH  CURVED  TOP. 


2  U3  U4  U5  U6  U7  U8 

FIG.  83.  —  HOWE  TRUSS,  INCLINED  TOP  CHORD. 


STRAINS  IN  FRAMED   OR  BUILT  GIRDERS.  131 


FIG.  84.  —  RAFTERS,  WITH  VERTICAL  TIE. 


FIG.  84*2.  —  SYSTEMS  OF  THE  SCHAFFHAUSEN  TRUSS. 
FORMULA  FOR  CLASS  X.     ft  =  o,  <£  =  90°. 

Method  of  Moments. 
If     =  M  -f-  h. 


. 

hr  +  i.         hr 

Hr  +  T   -5-  COS  «. 

-Hr. 


P  = 

U  = 

Y  = 

Z  = 


132 


MECHANICS  OF  THE   GIRDER. 


CLASS  XI.  —  TOP  CHORD  HORIZONTAL,     a  =  o.     STRUTS 

INCLINED.       TIES    VERTICAL.       <j>  =  90°. 


R  P2  Pa 


Pe 


U6 
FIG.  85.  —  SUSPENDED  Bow. 


FIG.  86.  —  FILLED  SPANDRELS. 


FORMULAE  FOR  CLASS  XI.     a  =  o,  <£  =  90 

Method  of  Moments. 
H     =  M  -*-  h. 


Mr 


ir  +  i 


U  = 
Y  = 
Z  = 


cos  ^. 


STRAIN'S  IN  FRAMED   OR  BUILT  GIRDERS.  133 


CLASS    XII.  —  BOTH    CHORDS   HORIZONTAL,     a  =  o, 
STRUTS  INCLINED.     TIES  VERTICAL.     </>  =  90°. 


=  o. 


R 


P5 


P6  P7 


U2  U3  U4,  U5  U6 

FIG.  87.  —  THE  HOWE  TRUSS. 


U8 


Y3 


U,  U2  U3  U*  U6  U6  U7  U8 

FIG.  88.  —  HOWE  TRUSS,  INCLINED  END  POSTS. 


FORMULAE  FOR  CLASS  XII.     a  =  o,  ft  =  o,  <£  =  90°. 


Method  of 


Moments. 


Moments  and  Shearing-Forces. 


H 


M  -T-  h. 
-5-  h. 


P 

U 

Y 


H  = 


U  =  -HW+L. 
Y   =  -S+I. 
Z    =  6*  -T-  sin  0. 


134 


MECHANICS  OF   THE    GIRDER. 


CHAPTER  V. 

MOMENTS    OF    RESISTANCE   OF  THE   INTERNAL   FORCES    OF  A  BEAM 
OR   GIRDER   HAVING   A   CONTINUOUS    WEB. 

SECTION  i. 

General  Formula  found  and  applied  to  Particular  Cross-Sections  of 
Beams  with  Continuous  Web. 

50.    The   mode   of    estimating   the   moment   of  resistance 

offered  by  the  cohe- 
sion of  the  particles 
of  the  material  com- 
posing a  beam,  we 
now  proceed  to  illus- 
trate. 

Let  ABCD9  Fig. 
89,  be  the  vertical 
longitudinal  central 
section  of  a  beam 
of  any  cross-section 
whatever,  under  the 
influence  of  given 
applied  forces  or 
pressures. 

It  is  required  to 
find  the  moment  of 
resistance  offered  by 


FIG.  89. 
beam  at  any  normal  section,  OTQ. 


the   material   of    the 


MOMENTS  OF  RESISTANCE   OF  INTERNAL   FORCES.     135 

Let  NS  be  the  intersection  of  the  neutral  surface  of  the 
beam  with  the  plane  of  the  paper.  The  neutral  surface  of  a 
beam  coincides  with  the  position  of  that  longitudinal  lamina 
which,  for  a  given  strain,  is  neither  compressed  nor  elongated. 

All  fibres  not  in  the  neutral  surface  are  assumed  to  be 
increased  or  diminished  in  length  by  a  quantity  in  direct  pro- 
portion to  their  distance  from  the  neutral  surface,  and  also  in 
direct  proportion  to  the  intensity  of  the  force  acting  on  the 
fibres. 

Let  /  =  the  force  acting  on  a  unit  of  area  of  any  given 
normal  section,  at  right  angles  to  the  section, 
and  at  the  unit's  distance  from  the  neutral  sur- 
face, either  above  or  below. 

dz  =  an  element  of  the  thickness  of  the  beam.^ 

dy  =  an  element  of  its  depth. 

y  =  the  distance  of  the  fibre  whose  area  is  dz  dy  from 
the  neutral  surface. 

Then  the  pressure  upon  the  area  dz  dy  is 

• 

fy  dy  dz, 
and  the  elementary  moment  due  this  stress  is 

dM  =  ffdydz; 
whence  the  total  moment  for  the  cross-section  is 

(159) 

which  expression  is  to  be  integrated  between  limits  depending 
on  the  form  of  the  given  cross-section  whose  centre  of  gravity 
may  be  taken  as  the  origin  of  co-ordinates. 


136 


MECHANICS  OF  THE   GIRDER. 


51.  We  will  now  apply  (159)  to  the  determination  of  the 
moment  of  resistance,  R,  for  various  cross-sections  occurring  in 
practice. 

Let  the  beam  have  a  rectangular  cross-section  of  the  breadth 
by  and  the  height  h,  as  in  Fig.  90. 


FIG.  90. 
Then  equation  (159)  becomes 


=  f 


y*dydz  = 


fdy, 


(160) 


where  /  is  the  unit  strain  at  the  unit's  distance  from  the  neu- 
tral surface,  and  B  =  \hf  =  the  unit  strain  at  the  distance  \h 
from  the  neutral  surface,  or  at  the  upper  and  lower  surfaces  of 
the  beam,  since  the  neutral  surface  is  here  assumed  to  be  in  its 
centre.  The  quantity  B  is  the  unit  strain  which,  at  the  instant 
of  rupture,  would  be  developed  at  the  upper  and  lower  surfaces 
of  a  beam  having  its  neutral  surface  midway  between  those 
outer  surfaces.  B  is  called  the  modulus  of  rupture,  or  the  ulti- 
mate unit  resistance  of  the  material  to  cross-breaking. 
A  table  giving  values  of  B  is  inserted  in  article  60. 


MOMENTS  OF  RESISTANCE   OF  INTERNAL   FORCES.     137 

52.   Beam  of   Hollow  Rectangular  Section,  or  Beam  of 
Equal  Flanges,  Fig.  91. 

b  b 


FIG.  91. 

Let  ^  =  height  of  beam. 

hi  =  height  of  cavity  or  web. 

b    =  breadth  of  beam  or  flange. 

#x  =  breadth  of  cavity. 

Then,  as  in  article  51,  we  shall  have,  for  the  whole  area 
b  X  h, 


and  for  the  area  of  the  cavity  #,  X  //»  or  2  X  J#x  X  ^,, 


Whence,  for  net  area  of  cross-section, 
R  = 


(161) 


where  ^  =  J/z/"  =  unit  strain  at  the  upper  and  lower  surfaces 
of  the  beam. 

If  the  beam  is  square  and  hollow,  so  that  h  =  b,  and  h*  = 
bu  we  have,  from  equation  (161), 


R  = 


(162) 


138 


MECHANICS  OF   THE   GIRDER. 


53.  Beam  composed  of  Two  Vertical  Plates  and  Two 
Horizontal  Channels. 


FIG.  92. 

\ 

Let  the  two  plates  and  the  two  channels,  Fig.  92,  have  equal 
cross-sections. 

b   =  entire  breadth  of  beam. 

h  =  entire  height  of  beam. 

bi  =  width  of  channel. 

b2  =  width  of  its  web. 

hT  =  distance  between  channels. 

h2  =  depth  of  channel  cavity. 

The  neutral  axis   (that  is,  the  line  of  intersection  of  the 
neutral  surface  with  the  normal  section)  is  here  central. 

Whence,  as  in  article  51,  we  have,  for  the  area  (b  —  b^  X  h, 


for  the  area  b2  X  (h  —  2//2), 


for  the  area  £,  X  hn 


MOMENTS   OF  RESISTANCE   OF  INTERNAL   FORCES.     139 


Whence  the  total  moment  of  resistance, 


R  =     [(j  -  b 

6/1 


-  w],    (163) 


where  B  =.  \hf  =  unit  strain  in  extreme  top  and  bottom  fibres. 
54.    Beam    composed    of    Two    Vertical    I-Beams   and 
Two  Equal  Horizontal  Plates. 


_    i 


FIG.  93. 

In  Fig.  93,  let  /£   =:  height  of  beam. 

hi  =  height  of  I-beams. 

//2  =  height  of  their  webs. 

b    =  width  of  plates. 

bi  =  width  between  beams. 

b2  =  width  of  cavities  of  beams. 

Then,  proceeding  as  in  the  last  article,  we  find 


R  =  £.(40  -  bji?  -  b 


(164) 


140 


MECHANICS  OF   THE   GIRDER. 


55-    Beam   composed    of    Two  Vertical    Channels    and 
Two  Horizontal  Plates. 

6 


A. 


FIG.  94. 

In  Fig.  94,  let  h  =  height  of  beam. 

^j  =  height  of  channels. 

h2  =  height  of  their  webs. 

b   =.  width  of  plates. 

bi  =  width  between  channels. 

b2  =  width  of  cavities. 
Then  we  have 


56.  Beam  with  but  One  Flange. 
'6 


N         i&i 

A 

£ 

fc,     Ti. 

B 

C 

(165) 


FIG.  95. 

Let  the  cross-section,  Fig.  95,  have  the  form  of  the  letter  T. 


MOMENTS  OF  RESISTANCE   OF  INTERNAL   FORCES.     141 

b  =  width  of  flange. 

h  =  whole'  height  of  beam. 

b  —  bl  =  thickness  of  web. 

h^          =  height  of  web. 

g  =  distance  of  neutral  axis  NA  from  any  line,  BC, 

parallel  to  NA,  in  the  plane  of  the  given  cross- 
section. 

ist,  To  find  «,  and  determine  the  position  of  the  neutral 
axis. 

Take  the  moment  of  the  area  of  the  section  about  BC  as  an 
axis,  and  divide  this  moment  by  the  area.  The  quotient  will 
be  the  value  of  «. 

Moment  of  flange  about  BC 

=  b(h  -  h,)  X  \(h  +  h,)  =  \b(h*  -  h*). 

Moment  of  web  about  BC  =  h*(b  —  dt)  X  JA,  =  \h?(b  -  h). 
Total  moment  of  area  about  BC  is 


Total  area  =  bh  —  bji*  ;  therefore 


" 


2d,  By  means  of  (159)  we  find, 
For  area  b  X  (h  —  s), 


for  area  (^  —  ^,)  X  «, 


I42  MECHANICS  OF  THE   GIRDER. 

for  area  b^  X  (/*i  —  «), 

*3  =  ^  f  y3y  =  i/&x(*i  -  «)3- 

t/o 

Whence  the  moment  of  resistance  due  to  the  net  cross-section 
is 


R  =      [*(*  -  8)3  +  (*  -  *«)«3  -  M*x  -  *)3]>    (I66> 

where  B  =  ef  =  unit  strain  at  the  extreme  edge  of  the  beam. 

In  a  similar  manner  may  all  other  beams  be  treated  whose 
cross-sections  are  composed  of  rectangles  having  two  sides  par- 
allel to  the  neutral  axis. 

57.  Solid  or  Hollow  Beam  of  Square  Cross-Section 
and  Diagonal  Vertical. 


FIG.  96. 

Let  b,  Fig.  96,  be  a  side  of  the  beam's  cross-section,  and  bl 
a  side  of  the  square  concavity  whose  centre  coincides  with  the 
beam's  centre.  Then  the  diagonals  are  b^2  and  b^2  ;  and 
(159)  becomes,  since  z  =  Jfl/2  —  y,  — 

For  solid  beam, 


R  = 

12 


where  ^  =  \fb2  =  intensity  of  stress  at  extreme  upper  or 
lower  edge  of  the  beam  whose  diagonal  is  vertical. 


MOMENTS  OF  RESISTANCE   OF  INTERNAL   FORCES.     143 

If  in  (160)  we  make  h  =  b,  then 

R  = 


where  B  =  \fb  =  intensity  of  stress  at  upper  or  lower  surface 
of  a  square  beam  whose  side  is  vertical. 

Hence,  although  the  identity  of  the  middle  members  of 
equations  (160)  and  (167)  shows  that  the  total  moment  of  resist- 
ance, R,  is  the  same  for  a  given  solid  square  beam  whether 
its  side  or  its  diagonal  be  vertical,  yet  the  extreme  fibres  for 
these  two  positions  of  the  beam  are  strained  in  the  ratio  of 
their  distances  from  the  neutral  axes;  that  is,  in  the  ratio 
of  i  to  ^2. 

If,  therefore,  B  expresses  the  ultimate  strength  of  the 
material,  when  in  equation  (167)  it  is  equal  to  J//H/2,  we  may 
evidently  give  to  B  the  same  extreme  value  in  equation  (160), 
and  thus  make  the  beam  V/2  times  stronger  when  its  side  is 
vertical  than  when  its  diagonal  is  vertical. 

Again,  for  the  vacant  square  whose  side  is  blt  since  z  = 
—  y,  we  have 


r  4*i/»          r- 

£=2/1    f(i4Vfl  - 

Jo 


And  therefore,  for  a  hollow  square  beam  with  diagonal  vertical, 

the  moment  of  resistance  is 

• 

X  =  J*B»  -  6'\  (168) 

12  b 


where  B  —  \fb^2  =  unit  strain  at  extreme  edge  of  beam  when 
the  diagonal  is  vertical. 


144 


MECHANICS  OF  THE   GIRDER. 


58.   Solid  or  Hollow  Beam  of  Circular  Cross-Section. 


FIG.  97. 

Let  r  =  radius  of  the  outer  circle,  Fig.  97,  and 

r,  =  radius  of  the  inner  circle. 
The  equation  of  the  outer  circle  is 


/.    z    =  (r*  - 
and  equation  (159)  becomes 


T— 

4 


(169) 


which  is  the  moment  of  resistance  for  a  solid  beam  of  circular 
cross-section  with  the  radius  r,  and  where  B  =  /r  =  the  unit 
strain  on  the  highest  and  lowest  fibres. 

If  the  beam  is  hollow,  the  inner  and  outer  circles  being 
concentric,  we  manifestly  have 


R  =  0.7854^'^-^, 
where  B  =  unit  strain  on  highest  and  lowest  fibres. 


(170) 


MOMENTS  OF  RESISTANCE   OF  INTERNAL  FORCES.     145 

59.  Beam  of  Elliptical  Cross-Section,  Solid  or  Hollow ; 
Longer  Axis  vertical ;  Axes  of  Outer  and  Inner  Ellipses 
coincident. 


FIG.  98. 


Let  k,  Fig.  98,  be  the  length  of  the  semi-transverse  axis  of 
the  outer  ellipse,  and  hl  that  of  the  inner  ellipse ;  b  =•  length 
of  semi-conjugate  axis  of  outer  ellipse,  and  £,  =  the  same  for 
the  inner  ellipse.  The  equation  of  the  outer  ellipse  is 

z2       y2 


Whence  (159)  becomes 

*"*lp 

7>  T 

•  •       jK.    =^    — 


0.7854^^, 


which  is  the  moment  of  resistance  for  the  solid  elliptical  beam, 
where  B  =  /if  =  unit  strain  on  highest  and  lowest  fibres. 


146  MECHANICS  OF   THE   GIRDER. 

Similarly,  for  the  area  of  the  inner  ellipse, 


and  therefore,  for  the  hollow  elliptical'  beam,  the  moment  of 
resistance  is 

R  =  0.7854***'  ~  W,  (i  70 

where  B  =  hf  =  unit  strain  on  highest  and  lowest  fibres. 

60.  These  illustrations  may  suffice  for  girders  of  continuous 
web. 

We  close  this  section  with  a  table  giving  the  limiting  value 
of  B,  in  pounds  avoirdupois  to  the  square  inch,  for  the  ordinary 
materials  used  in  beams  ;  that  is,  the  values  of  B  in  this  table 
are  values  which  cannot  be  exceeded  in  the  equations  of  this 
section,  and  represent  the  ultimate  resistance  of  the  material 
to  cross-breaking. 

It  should,  however,  be  borne  in  mind,  that  B  may  not  repre- 
sent the  actual  unit  strain  which  the  material  is  capable  of 
resisting,  either  in  tension  or  compression  ;  but  that  it,  in  gen- 
eral, has  some  mean  value  between  the  ultimate  resistance  of 
material  to  crushing,  and  the  ultimate  resistance  of  the  same 
material  to  tearing  by  direct  pull. 

Continuing  the  suppositions  made  in  article  50,  we  may  find 
the  relation  existing  among  these  three  ultimate  unit  strains  in 
rectangular  beams  as  follows  :  — 

Let  us  take 

h   —  depth  of  beam. 

b    =  breadth  of  beam. 

/    =  length  of  beam. 

x  =  distance  of  the  neutral  surface  from  the  compressed 
side  of  the  beam. 


MOMENTS   OF  RESISTANCE   OF  INTERNAL   FORCES.     147 

C  —  ultimate  resistance  of  the  material  to  crushing  by 
direct  thrust,  in  pounds,  per  square  inch. 

T  =  ultimate  resistance  of  the  material  to  extension,  in 
pounds,  per  square  inch. 

B  =  the  unit  strain  which,  at  the  instant  of  rupture,  would 
be  developed  at  the  upper  and  lower  surfaces  of  a 
beam  having  its  neutral  surface  midway  between 
these  outer  surfaces  ;  that  is,  B  =  the  modulus  of 
rupture,  also  in  pounds  per  square  inch. 

Then,  using  (159),  we  find  for  the  compressed  part  of  any 
cross-section,  moment  of  internal  forces, 


(173) 
and  for  the  extended  part  of  the  same  cross-section, 


But,  if  the  neutral  axis  bisected  the  given  cross-section,  we 
should  have  moment  of  internal  forces  on  either  side  of  this 
axis, 

R  =  \Bb(&hy.  (175) 


Now,  since  experimental  researches  show,  that,  for  many 
materials  used  in  construction,  these  three  expressions  are 
nearly  equal  to  one  another,  we  have,  approximately 


\Cb&  =  \Tb(h  -  *Y  =  \Bb(\K)*;  (ll6) 

whence 

JT  r=9      (177) 
(178) 


C  ^T  \/C  +  \/T 

B 


148  MECHANICS  OF  THE   GIRDER. 


(179) 


(180) 


When,  therefore,  any  two  of  the  three  quantities,  C,  T,  and  B, 
are  given,  the  third  may  be  found,  and  also  the  position  of  the 
neutral  surface. 

It  is  probable,  that  after  the  elastic  limit  of  the  material  is 
passed,  and  rupture  is  about  to  take  place,  the  expressions  in 
(176)  do  not  represent  the  actual  moments,  but  are  similar 
functions  of  C,  T,  and  B,  and  are  proportional  to  the  forces  then 
developed.  For  within  the  elastic  limits  the  forces  are 

\CJ>x  =  \TJ>(h  -  x)  =  \BJ>(&K),  (181) 

C,,  TI9  and  BT  being  the  limited  unit  strains  at  the  surfaces. 
But  when  the  strain  on  the  extreme  fibres  passes  the  elastic 
limit,  and  the  fibres  expand  or  contract  more  rapidly  than  the 
strain  increases,  then  an  increment  is  given  to  all  the  previous 
inner  strains  proportional  to  their  distances  from  the  neutral 
surface,  which  is  equivalent  to  introducing  a  factor  of  the  form 
kx  into  the  expressions  (181),  whereby,  at  the  instant  of  rup- 
ture, they  become 

\Cbkx*  =  \Tbk(h  -  x}2  =  \Bbk(\hY,          (182) 

which  is  identical  with  (176). 

Experiments  indicate,  that  in  the  case  of  cast-iron,  owing 
to  the  superior  hardness  of  the  outer  over  the  inner  portions  of 
the  metal,  an  increment  should  be  given  to  B  in  (180)  equal  to 
one-ninth  of  itself. 


MOMENTS  OF  RESISTANCE   OF  INTERNAL   FORCES.     149 


TABLE  II. 


MATERIAL. 

Authority. 

Ultimate  Resistance,  in  Lbs.,  per 
Square  Inch,  to 

Modulus  of 
Elasticity, 
E. 

Com- 
pression, 
C. 

Tension, 
T. 

Cross- 
Breaking, 
B. 

Cast-Iron. 

Stoney. 

« 
« 
« 

Rankine. 
Thurston. 

Stoney. 

« 

it 

Rankine. 
« 

« 
« 

Lovett. 
Stoney. 

105945  H. 
86284  H. 

II2000 
87429 
127323 

40320 
36000 
40000 

225568  F. 

16720  H. 
15298  H. 

16500 
20500 
344°7 

57555  K. 
60000 
70000 
51000 

357oo 
28600 
64000 
70000 

100000 

90000 

50915 

83391  F. 
71658  K. 

37695  H.F. 

45696  C. 
30240  C. 
26880  C. 
38304  C. 

45965  c. 
38250 
45760 
67035 

51341  c. 

74995  C. 
30240  C. 
70291  C. 
43814  C. 
61824 
53760 
51475 
52567  C. 

42000 

128083  K. 
115181  K. 

12090000 

17000000 

"450754 
15968254 

24000000 
29000000 

25300000 
15000000 

27311111 
31000000 

Means  of  16  irons     .     „     .     . 
Bars  not  >  i  inch  wide  .     .     . 
Bars  3  inches  wide     .... 
Bars,  small  round      .... 

Wrought-Iron. 

Bars,  previously  strained   .     . 
Bars,  new  round   
Boiler  tubes,  welded  .... 
Circular  tubes,  riveted  .     .     . 
Rolled  I-beams,  about   .     .     . 
T-iron,  flange  up  about  .     .     . 
T-iron,  flange  down  about  .     . 
Average  
Bars  and  bolts  .     .         ... 

Plates                

Plates,  double-riveted    .     .     . 
Plates,  single-riveted      .     .     . 

Wire   

Wire 

Mean  of  113  tests      .... 
Mean  of  27  tests   ..... 

Steel. 
Bessemer,  hammered     . 
Bessemer,  rolled   

ISO 


MECHANICS  OF  THE   GIRDER. 


TABLE  II.  — Continued. 


MATERIAL. 

Authority. 

Ultimate  Resistance,  in  Lbs.,  per 
Square  Inch,  to 

Modulus  of 
Elasticity, 
E. 

Com- 
pression, 
C. 

Tension, 
T. 

Cross- 
Breaking, 
B. 

Steel  (continued). 

Crucible,  hammered  .... 
Crucible,  rolled      ..... 
Cast,  not  hardened    .... 
Cast,  low  temper  ..... 

Stoney. 

M 

« 
« 

Rankine. 
<« 

Stoney. 
Rankine. 
Stoney. 

M 

«( 

Rankine. 
Stoney. 

Rankine. 
Stoney. 

Rankine. 
Haswell. 
Stoney. 

« 

Rankine. 
Stoney. 

198944  Wd. 
354544  Wd. 
391985  Wd. 
372598  Wd. 

6831  H. 
9363  H. 
9000 
11500 
9363  H. 

11663  H. 
6402  H. 

10300 

5863  H. 
5860 

5350 

6586  H. 
7293  H. 
10300 
10331  H. 

85546  K. 
68589  K. 

1  00000 

130000 
80000 

13900  Mu. 
16700  Bv. 
170006. 
9360 
115006. 
17300  Mu. 

15000  Bv. 

20000  B. 
20000 

11400  Bv. 
11400 
13300  Ro. 
10500  Bv. 
12100  Bv. 
11500 

6000  Mu. 
12900  Bv. 

14000 
14400  Bv. 

147840  K. 
118272  K. 

"7344 

12156  B. 
13000 
10500 
9336  B. 

12366  B.D. 
11568  B. 
14670  T. 

4596  D. 
8958  D. 
7400 

10660 
9372  B. 

7850 

4692  B.D. 
11820  D.N. 

29000000 
42000000 

1600000 
1350000 

1645000 

486000 
1140000 

1020000 

Cast,  mean  temper    .... 
Cast,  high  temper      .... 
Bars                  .... 

Bars                  

Plates,  average     

Average  

Wood. 
Alder  

Ash     

Ash     

Beech  

Beech  

Beech  

Birch,  American   

Box     

Box 

Cedar,  American  white  .     .     . 
Cedar  of  Lebanon     .... 
Cedar  of  Lebanon     .... 
Chestnut,  Spanish     .... 

Chestnut,  horse     .     .     . 

Deal,  Christiana  .     . 

Deal,  red      

Deal,  white 

Elm     

Elm     

Elm,  English    ... 

Elm,  Canada  Rock    .... 

MOMENTS   OF  RESISTANCE   OF  INTERNAL   FORCES.     151 


TABLE  II.  —  Continued. 


MATERIAL. 

Authority. 

Ultimate  Resistance,  in  Lbs.,  per 
Square  Inch,  to 

Modulus  of 
Elasticity, 
E. 

Com- 
pression, 
C. 

Tension, 
T. 

Cross- 
Breaking, 
B. 

Wood  (continued)  . 
Fir,  spruce  
Fir,  spruce,  American  black   . 

Stoney. 
Rankine. 

Stoney. 

N 
« 

Rankine. 

Stoney. 
Haswell. 
Rankine. 

Stoney. 

« 

Rankine. 
Haswell. 
Rankine. 

Stoney. 
<« 
a 
a 

6819  H. 

5375 
6200 
5400 

5570 

5568  H. 

9900 

9"3 

8200 
8198  H. 

8150 

7700 

10000 

6000 
10058  H. 

5982  H. 

7518  H. 
6790  H. 

I2OOO  B. 
I2OOO 
I4OOO 

12400 

9OOO 
IOOOO 

10220  Ro. 

8900  Bv. 

11800  Bv. 
11800 
16000 
20100  Mu. 

8000 
21800 
8000  B. 
16500  Bv. 
17400  Bv. 
10600 

IOOOO 

19800 
10250 

IOOOO  B. 

19800  Bv. 
13950  Ro. 

7650  Mu. 

8076  M. 
6216  D. 
7392  B. 
7100 
954° 

9900 
12300 
5000 

IOOOO 

6852  D. 
12774  D.N. 
8790  D. 
8010  B.D. 

5466  D. 
12078  N. 

I2OOO 
II2OO 

20580  B. 

7600 
II500 

10314  M.N. 
10164  D. 

8700 
13600 
10600 
10164  B.D. 

8898  M. 

IOI22D.N. 

10458  B.D 
9162  B.D. 
10362  B.D. 
7374  D.N. 

1460000 
1900000 

1400000 
1800000 
900000 
1360000 

1255000 

1200000 
1750000 
2I5OOOO 

Fir,  yellow  pine,  American     . 
Fir,  spruce  

Fir  larch 

Hemlock 

Hickory,  American  .... 
Hickory,  bitter-nut  ..... 
Larch 

Larch                         .... 

Larch,  American  
Lignum-vitae     

Lignum-vitae     

Locust 

Locust                             .    . 

Locust               .              ... 

Maple      

Maple      

Maple      
Oak,  European     . 

Oak,  European      .... 

Oak,  American  red   .... 
Oak,  English    
Oak,  English    
Oak,  French 

Oak,  Quebec 

Oak,  American  red    .... 
Oak,  American  white     .     .     . 
Pine,  American  red  .... 
Pine,  American  pitch     .     .     . 
Pine,  American  white     .     . 

152 


MECHANICS  OF  THE   GIRDER. 


TABLE  II.— Continued. 


MATERIAL. 

Authority. 

Ultimate  Resistance,  in  Lbs.,  per 
Square  Inch,  to 

Modulus  of 
Elasticity, 
E. 

Com- 
pression, 
C. 

Tension, 
T. 

Cross- 
Breaking, 
B. 

Wood  (continued). 

Pine,  American  yellow  .    .    . 
Pine,  Norway  

Stoney. 

« 

Rankine. 
Stoney. 

Rankine. 

M 

Stoney. 
« 
« 
Rankine. 

Stoney. 

Rankine. 
« 
« 
« 
Stoney. 

Haswell. 
« 

Stoney. 
,   « 

Rankine. 

Moseley. 
Stoney. 

Rankine. 
Stoney. 

5445  H. 
7082  H. 

I2IOI  H. 
12000 

7227  H. 
6128  H. 

3i73  Wi. 
i344oWi. 
55°o 

IIOOO 

4000 

4500 

3050  F. 
18043  Wi. 

3216  Re. 
20160  Wi. 
5500 

2185 
7884 
55oo 

2200 
17344 

14300  Bv. 
7287  Bv. 
13000 
13000  Bv. 
15000  B. 
15000 

8130  Mu. 
7800  Bv. 
14000  Bv. 

670 
2800 
55i  H. 
722  Bu. 

1054  Bu. 
1261  Bu. 

9600 
12800 

7110  B.D. 

9600 
12648  B.M. 

I2OOO 
igOOO 
14980 

6600 

456  Wi. 
2442  Wi. 

1698  Wi. 
2484  Wi. 

1252  H. 
2697  H. 
2664 
2010  Re. 
5142  Re. 
2360 

IIOO 

5000 

7370 

1040000 
2400000 

13000000 
16000000 

Pine,  Norway        

Sycamore          

Teak  

Teak,  Indian    

Teak,  Indian    

Teak  African 

Walnut    

Walnut 

Willow 

Willow 

Stone. 

Granite 

Limestone 

Limestone             . 

Limestone    . 

Limestone    . 

Limestone    . 

Limestone    .... 

Marble    

Marble    >   .     .     .    . 

Marble    

Marble,  white  .... 

Marble,  black  

Marble,  black  

Sandstone    

Slate   

Slate   

Slate   

MOMENTS  OF  RESISTANCE   OF  INTERNAL   FORCES.     153 


TABLE  II.  —  Concluded. 


MATERIAL. 

Authority. 

Ultimate  Resistance,  in  Lbs.,  per 
Square  Inch,  to 

Modulus  of 
Elasticity, 
E. 

Com- 
pression, 
C. 

Tension, 
T. 

Cross- 
Breaking, 
B. 

Bricks,  etc. 
Pale  red  

Stoney. 

Rankine. 
« 

Stoney. 

« 

ft 

562  Re. 
808  Re. 
1717  Re. 
2240  Gr. 

618  Ro. 
5984  Gr. 

280 
300 
51 
358  Gr. 
71  Ro. 

546  Gr. 
604  Gr. 
632  Gr. 
627  Gr. 
666  Gr. 
709  Gr. 

- 

•~ 

Red 

Fire 

Gault  clay                       . 

Lime  mortar,  average    .     .     . 
Portland  cement    

Plaster  of  Paris     

ROMAN  CEMENT:  — 

The  value  of  B,  the  modulus  of  rupture  in  Table  II.,  is  that 
due  to  a  rectangular  cross-section,  unless  otherwise  specified. 

The  works  from  which  this  Table  is  made  up  are  the  fol- 
lowing well-known  authorities :  — 

ist,  "A  Manual  of  Civil  Engineering,"  by  William  John 
Macquorn  Rankine. 

2d,  "  The  Theory  of  Strains  in  Girders  and  Similar  Struc- 
tures," by  Bindon  B.  Stoney. 

3d,  "  The  Mechanical  Principles  of  Engineering  and  Archi- 
tecture," by  Henry  Moseley. 

4th,  "  Engineers'  and  Mechanics'  Pocket-Book,"  by  Charles 
H.  Haswell. 

5th,  "  Report  on  the  Progress  of  Work,  etc.,  of  the  Cincin- 
nati Southern  Railway,"  by  Thomas  D.  Lovett. 


154  MECHANICS  OF  THE   GIRDER. 

6th,  "  Report  on  Tests  of  Salisbury  Cast-Iron,"  in  "  Railroad 
Gazette"  of  Nov.  30,  1877,  by  Robert  H.  Thurston. 

Names  of  the  experimenters  cited  are  thus  abbreviated  :  viz., 
H.,  Hodgkinson  ;  F.,  Fairbairn ;  Bv.,  Bevan ;  Bu.,  Buchanan ; 
B.,  Barlow ;  D.,  Denison ;  N.,  Nelson ;  M.,  Moore ;  K.,  Kir- 
kaldy ;  Ro.,  Rondelet ;  Re.,  Rennie ;  C,  Clark ;  Gr.,  Grant ; 
Wi.,  Wilkinson ;  Wd.,Wade;  Mu.,  Musschenbroek;  T.,  Trickett. 


SECTION  2. 

Moment  of  Inertia  and  Radius  of  Gyration  of  a  Given  Cross-Section. 
61.  In  equation  (159)  the  factor 

fffdydz^I  (183) 

is  called  the  moment  of  inertia  of  the  surface  of  the  cross-sec- 
tion, relatively  to  the  axis  of  z,  the  factor  being  analogous  to 
the  real  moment  of  inertia  of  a  material  plate  whose  thickness 
is  unity. 

The  moment  of  inertia  divided  by  the  area  of  the  section 
gives  the  square  of  the  radius  of  gyration,  which  we  will  call 
r2. 

We  then  have,  if  5  is  that  area, 

Square  of  radius  of  gyration  =  r2  =  -~  =     rr  '  —  .       (184) 


62.  From  the  moments  of  resistance  already  found,  equa- 
tions (160)  to  (171),  and  from  similar  applications  of  (183),  we 
derive  values  of  /and  of  r2  as  given  below  in  Table  III.,  where 
the  axes  of  gyration  are  assumed  to  pass  through  the  centre  of 
gravity  of  the  cross-section. 


MOMENTS  OF  RESISTANCE   OF  INTERNAL   FORCES.     155 


TABLE    III. 


FORM  OF  CROSS-SECTION. 

Moment  of  Inertia 
of  Section, 

Square  of  Radius 
of  Gyration, 
r2 

i.  Rectangle  (Fig.  90). 
About  least  axis  b     

Max. 
Min. 

Max. 
Min. 

| 

I 

•Yqbh* 
\$b  h 

-fa(bh?  —  Mi3) 

bh*  —  Mi3 

About  greater  axis  h     .    .    .    . 
2.  Square. 

3.  Hollow  Rectangle  (Fig.  91). 

About  greater  axis  h     .    .    .     . 
4.  Hollow  Square. 

\2(bh  —  l>iAi) 

^(h  —  h2)b3 

\2(bh  —  MX) 
•&(hz  +  hf). 

b2A 

5.  I-Section  (Fig.  91). 
About  vertical  axis  h    .    .     .     . 

About  horizontal  axis  b  .    .    .    . 

A  •=.  area  of  flanges. 
B  =  area  of  web. 

6.  Plates  and  Channels  (Fig.  92). 
About  axis  6,  normal  to  plates  . 

About  axis  l>,  parallel  to  plates  ) 
(Fig.  04.)    .                                 j 

12(A  +  B) 

12(A  +  B) 
7-T61. 

/^•S1. 
7-=-^. 

7.  Plates  and  I-Beams  (Fig.  93). 
About  axis  b,  parallel  to  plates  . 

About  axis  /;,  normal  to  plates  . 

i56 


MECHANICS  OF   THE   GIRDER. 


TABLE   III.  —  Concluded. 


FORM  OF  CROSS-SECTION. 

Moment  of  Inertia 
of  Section, 
7. 

Square  of  Radius 
of  Gyration, 
r2. 

8.  T-Section,  erect  (Fig.  95). 
About  axis  b,  parallel  to  flange 
bhz  —  Mi2    _  hej-ht  of 

I 

\b(h  -  e)3 

I  —  S 

2(bh  —  <MX) 
neutral  axis. 

About  axis  h>  normal  to  flange  . 
9.  Angle    Iron;    equal  ribs  3,  thick 

1 

Min. 

j££ 

• 
I  '-r  S. 

Of  unequal  ribs  h  and  b     .    .    . 

Min 

Phz(b  +  h}t 

Ph* 

10.  Channel    Iron;    h  =  depth    of 
flanges  -f  £  thickness  of  web, 
A  =  area  of  flanges,  B  •=. 
area  of  web  

Min. 

4*4) 

^(^,+^D- 

n.  Star    Iron,    or    cross    of    equal 
arms  h      

Min 

JLji 

12.  Ellipse  (Fig.  98). 
About  minor  axis  2b      .... 
About  major  axis  2h      .... 

13.  Circle  ;  radius  h  (Fig.  97)     .     .     . 

14.  Hollow  Ellipse  (Fig.  98). 
About  minor  axis  2b 

Max. 
Min. 

if 

About  major  axis  2h 

Min 

4 

15.  Hollow  Circle  (Fig.  97). 
Radii,  h,  h*  

4 

4 

#w 

DEFLECTION,  END  MOMENTS,  ETC.,  FOUND.  157 


CHAPTER    VI. 

DEFLECTION,    END    MOMENTS,    AND   POINTS    OF   CONTRARY 
FLEXURE    FOUND. CAMBER. 

SECTION  i. 
Deflection  of  the  Semi-Beam  having  a  Uniform  Cross-Section. 

63.  Equation  of  the  Elastic  Curve  as  applied  to  a  Beam 
or  Pillar.  —  Let  N,  Fig.  89,  be  the  origin,  and  x  and  y  the 
current  co-ordinates,  of  the  neutral  line  NTS  of  any  beam  or 
column  under  a  given  load ;  TU  =  a  unit  of  the  length  of  the 
beam  ;  VT  —  VU  =  the  radius  of  curvature  at  any  point  = 


P=" ^T^=-^  ('85) 


when  the  deflection  of  the  beam  or  pillar  is  small  compared 
with  its  length.  (See  Differential  Calculus.)  PP^  =  incre- 
ment of  unit  on  extended  side  due  to  flexure ;  RR^  =  decre- 
ment of  unit  on  compressed  side  due  to  flexure ;  a  =  the 
angle  included  between  the  tangent  to  the  curve  at  any  point 

and  the  axis  of  x,  so  that  tan  a  =.  -^-. 

dx          j 

Suppose  the  unit  strain  required  to  extend  a  unit  of  length 
by  the  space  PP^  to  be  Tiy  and  that  required  to  compress  a 


158  MECHANICS  OF   THE   GIRDER. 

unit  of  length  by  the  space  RR^  to  be  Cu  and  that  required  to 
extend  or  compress  a  unit  of  length  by  its  own  length,  TU,  to 
be  E  =  the  modulus  of  transverse  elasticity  ;  and  put  Cx  +  Tt 
=  2Bi  =  total  unit  strain  at  surfaces. 

We  then  have,  if  h  is  the  depth  of  the  beam,  and  if  the  dis- 
placing forces  E  and  zB^  are  proportional  to  the  displacements 
they  cause, 

P,P  ^R,R  =  TU 
PU       RU    "  UV9 


f  - 


Multiplying  (186)  by  /  =  moment  of  inertia  of  cross-section, 
we  find 


which  is  the  moment  of  resistance  of  the  cross-section,  since 
Bl  -T-  \h  =  ft  and  /  -=ffy2dydz  of  equation  (159). 

If,  therefore,  we  put  —EI—^  equal  to  the  moment  at  the 

section  due  the  external  forces  acting  on  a  beam  or  pillar  of 
uniform  cross-section,  and  perform  two  successive  integrations, 
we  shall  have  an  equation  in  which  y  is  the  deflection  of  the 
neutral  line  at  the  distance  x  from  the  origin  of  co-ordinates. 

64.  Deflection  of  the  Semi-Beam  under  One  Weight. — 
Let  L}  Fig.  8,  the  point  where  the  neutral  line  of  the  semi- 
beam  meets  the  wall,  be  the  origin  of  co-ordinates,  and  call  x 
positive  to  the  left,  and  y  positive  downward,  in  accordance  with 
the  notation  in  article  14. 

Take  semi-beam  of  length  /,  with  concentrated  load  W  at 
distance  a'  from  fixed  end. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  159 

From  equation  (18), 

Mx  =  -  W(a'  -  x)  =  -EI^Z 
by  (187), 


dx2 
Integrating,  with  the   condition  that  -2-  =  o  when  x  —  o, 

6?Jv 

we  have 

EI&  =  wit 

'  dx  \ 

Integrating   again,  with   the   condition   that  y  =  o  when 
x—  o, 

/.     ^"Ty  =   i 


Deflection  at  any  point,  x>  is 

M7 

(188) 


And  when  x  =  ^'  =  /,  we  have 

Maximum  deflection  Z>  =  -,  (189) 


where  the  proper  values  of  E  and  I  are  to  be  taken  from  Tables 
II.  and  III.,  according  to  the  material  used,  and  the  form  of 
cross-section. 

65.  Semi-Girder,  Length  /,  Uniform  Load  w  per  Unit.  — 

From  equations  (23)    and  (187), 


=  %w(l  -  *)*. 


i6o 


MECHANICS  OF  THE   GIRDER. 


When       =  o,*  =  o; 
dx 


dx 

When  y  =  o,  x  —  o ; 
.-.    Ely  = 

Deflection  anywhere, 


Maximum  deflection, 


(I90) 


(191) 


66.  Semi-Girder,  Partial  Uniform  Load,  «/,  on  each  Unit 
of  Length,  b,  at  Distance  a  from  Fixed  End.  Fig.  8.  — 
When  x  —  a,  or  x  <  a,  equations  (29)  and  (187)  give 


-  x) 


dy 

-f.  =  o  when  ^r  =  o, 

dx 


y  =  o  when  ^r  =  o, 


DEFLECTION-,   END  MOMENTS,   ETC.,   FOUND.  l6l 

Deflection  x  not  >  a, 

*  (I92) 


Again,  -^  —  tan  a  when  x  =  <z, 
dx 


=:  o  when  jr  =  o, 
^/(j,  _  x  tan  a)  =  o/3  1  (\b  +  «)(-  -  ^  -  (—  - 

Let  7  =  jz  when  ^r  =  a, 

.'.     Ela  tan  a  -  Ely,  =  Ja/^2^  + 
From  (192), 


/.     ^"7  tan  a  =  \w'ab(b  +  «). 

Also,  when  ,r  is  not  <  #  nor  >  (a  +  3),  we  have,  from  (26) 
and  (187), 


=  tan  a  when  x  =  a, 


£! I 

) 


1  62  MECHANICS  OF   THE   GIRDER. 

y  •=.  y^  when  x  =  a, 

.-.     EI(y  -  yt)  -  (x  —  a)  iz.na.EI 


-  a}\. 
) 


4 
Whence,  after  eliminating  yt  and  tan  a,  we  find  the  deflection 


+  6(a  +  bY(o?  -  a2)  -  ^(x  -  a) 

+  6a*b2  +  803J],  (193) 

And  if  ^  =  a  +  ^,  we  have 
Maximum  deflection  D  —    ^,[3(0  +  <^)4  —  3^4  —  4«3^].        (194) 


67.  If  it  be  required  to  find  the  total  deflection  of  a  semi- 
beam  at  its  free  extremity  when  it  supports  partial  uniform  or 
concentrated  loads  not  reaching  that  extremity,  we  may  proceed 
as  follows  :  — 

Find  the  deflection  at  the  free  end  due  the  beam's  own 
weight,  Iw  ;  then  find  the  deflection,  D,  and  the  inclination,  a, 
of  the  beam  at  any  point  bearing  a  concentrated  load,  W,  or  at 
the  outer  end  of  any  partial  uniform  load,  bid  ;  using  tan  a, 
compute  the  end  deflection  due  W  or  bw'  by  the  formula 


+  (/  —  a  —  £)tan  a  for  bwr, 
£>2  =  Dw  +  (/  -  </)tan  a  for  W. 

These  deflections  added  to  that  due  the  beam's  own  weight, 
will  give  the  total  deflection  at  the  free  end  of  the  semi-beam. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  163 

EXAMPLE.  —  Suppose  the  semi-beam,  Fig.  8,  to  be  of  oak, 
weighing  54  pounds  to  the  cubic  foot,  and  to  have  a  rectangular 
cross-section  I  foot  deep  and  \  foot  wide,  so  that  weight  of 
beam  per  foot  of  length  =  w  —  27  pounds  ;  length  ==  15  feet, 
b  =  4  feet,  loaded  with  wr  =  100  pounds  per  foot,  beginning 
a  =  5  feet  from  the  fixed  end  of  beam  ;  W  =  100  pounds, 
placed  a'  —  1  1  feet  from  fixed  end  ;  E  =  2,000,000  pounds  per 
square  inch  =  288,000,000  pounds  per  square  foot. 

From  Table  III.,  /  =  -^bk2  =  TV  X  J  X  i2  =  £$, 


.*.       El  =    I2OOOOOO. 

Deflection  due  beam's  own  weight  is,  by  (191), 


27  x  i54  x  12 

8  x  12000000   =  °'I'°86  mch' 


From  (194), 


Differentiating  (193),  we  have 

dy 

-*-  =  tan  «! 

=  [4-#3  —  12(0  +  b)x*  -f-  12(0  -|-  b}2x  —  4#3] 

IOQ  X  2416 
24  X   12000000 
when  x  =  a  +  b  =  g  ; 


...     (/  -  g  -  j)tan«.  =  6  x  IO°  x  ^6  x  I2  =  0.0604  inch, 

24    X    12000000 

.*.    Dj.  =  0.06587  -f  0.06040  =  0.12627  inch  at  end  due  M. 


1 64  MECHANICS  OF  THE   GIRDER. 

From  (189), 

DW  =  -^ — —    =  0.1125 

3  x  12000000 

Differentiating  (188),  we  find 

dy  W I  ,         x2\      100  x  4  X  121 

-^-  =  tan«2  =  ——lax )=  2 

ax  •£'1\  2  /  12000000 

when  x  =  a'  —  1 1 ; 

(J       ,w  4  x  100  x  \  X  121  x  12 

(/—  a  )tana2  = ^ =  0.0242 

12000000 

.-.     D2  =  0.1125  +  0.0242  =  0.1367  inch  at  end  due  W. 
Therefore  total  deflection  at  free  end  is 
D  =  Dw  +  Dl  +  D2  =  0.17086  -f-  0.12627  -f-  0.1367  =  0.43383  inch. 

68.  If  we  have  any  number,  r,  equal  weights,  W,  placed  at 
equal  intervals,  -,  along  the  semi-girder,  the  first  weight  being 

a  full  interval  from  the  fixed  end,  and  the  nth  or  last  weight 
being  at  the  free  end  when  the  beam  is  fully  loaded,  then  the 
total  deflection  at  the  free  end  due  to  the  first  r  equal  weights 
may  be  found  as  follows  :  — 

In  equation  (188)  put  x  =  af  =  -  ;  then 

n 


which  is  the  deflection  at  the  rth  weight  due  to  that  weight 
alone. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  165 

But  the  deflection  at  the  free  end  due  the  rth  weight  is 
greater  than  the  deflection  at  the  rth  weight  by  the  product  of 
the  tangent  of  the  slope  a  of  the  beam  at  and  beyond  the  rth 
weight,  multiplied  by  the  horizontal  distance  between  this 
weight  and  the  free  end  ;  and  this  horizontal  distance  is  prac- 
tically equal  to  f  I }  =  the  part  of  the  beam's  length  be- 
yond the  rth  weight. 

By  differentiating  (188),  we  find 


if  x  =  af  =  -,  and 
n 


Add  value  of  y  in  (195)  =  -=r(-J  —  >  and  we  have  deflection  at 

±L1  \1l  /    3 

free  end  due  the  rth  weight, 


<-> 

Hence  the  end  deflection  due  the  first  r  weights  is 


+  i)  -r^r+i)'].     (197) 
When  r  =  ;/,  (197)  becomes 

/>  =  .»ZL(«  +  'Kv  +  Ot  (,98) 

24^7  n 

which  is  the  deflection  at  the  free  end  due  n  equal  weights  at 

equal  intervals,  -. 
n 


166  MECHANICS  OF   THE   GIRDER. 

69.  If  the  weight  at  the  free  end  is  but  \W,  while  every 
other  weight  is  W,  as  is  generally  the  case  with  a  uniform  load 
of  panel  weights,  we  must  diminish  the  deflection  last  found  by 
the  deflection  due  J  W  at  the  free  end,  which,  according  to 
(189),  is 


This  quantity  taken  from  the  value  of  D  in  (198)  leaves 

W>    j(,,+    I)(y,+    i)  1 

S"~       ""  "4'          (I99) 


which  is  the  deflection  at  the  free  end  of  a  semi-girder  of  uni- 
form cross-section  when  the  load  is  distributed  in  equal  panel 
weights  ;  there  being  but  the  half  panel  weight  at  the  free 
end. 

70.  To  find  the  Deflection  at  the  rth  Interval  due  to  all 
the  n  Equally  Distributed  Equal  Weights,  W. 

For  the  first  r  weights,  (198)  applies  if  we  put  r  for  n,  and 
/,  for  /;  /j  being  the  distance  of  the  rih  weight  from  the  fixed 
end.  That  is, 

Wl?  (r+  i)(3r+  i) 


For  the  (n  —  r)  weights  beyond  the  rth,  we  have,  from  (188), 
since  x  now  becomes  /„ 


y  = 


the  deflection  at  the  rth  weight  due  to  any  one  weight  at  the 
distance  a'  from  the  fixed  end, 


DEFLECTION,  END   MOMENTS,   ETC.,   FOUND. 


Therefore,  by  summing  (201),  we  find 


_{] 
~  EI2 


the  deflection  at  the  rth  weight  due  all  the  weights  beyond. 
Adding  (200)  and  (202),  there  results 


D  =  -r(6«2  +  r*  -  4rn  -  2r  +  6n  +  i),       (203) 


which  is  the  deflection  at  the  rth  weight  due  to  all  the  n  given 
weights,  W. 

71.   If  the  weight  at  the  free  end  is  the  m^  part  of   Wt 

instead  of   Wy  the  deflection  due  —  at  the  rth  point  of  divis- 

m 

ion  is,  by  putting  a'  —  /,  and  /z  =  — ,  in  (201), 


=  _  _ 

"  2  i     " 


mEI2n2        6ni          2EIn  m 


Subtracting  this  value  of  y  from  the  deflection  in  (203),  we 
have,  finally, 

1M  +  r*-4rn-2r  +  6n  +  i-  I2n  "  *r\ 
\  m        I 

-  r2  +  i)       (204) 


if  m  =  2 ;  st>  that  (204)  is  the  deflection  at  the  rth  panel  point 
due  to  a  full  load  of  equal  panel  weights. 


1 68  MECHANICS  OF   THE   GIRDER. 

EXAMPLE.  —  Take  the  oak  semi-beam  of  the  example  in 
article  67,  and  suppose  it  loaded  with  n  —  5  equal  weights,  of 
100  pounds  each,  at  intervals  of  3  feet.  El  =  12,000,000. 

Then  the  deflection  at  the  free  end  due  the  5  weights  is, 
by  equation  (198), 

_        IPO  X  is3  (5  +  i)(3  X  5  +  i) 

24  x  12000000  5 

=  0.0225  foot  =  0.27  inch; 

to   which   add  0.17086,   the   deflection   due   the   beam's   own 
weight,  for  the  total  deflection  at  the  free  end  =  0.44086  inch. 
If  the  fifth  or  end  weight  is  but  J  X  100  =  50  pounds, 
then,  by  (199), 

IPO  x  153     |  (5  +  i)(3  x  5  +  i)         ) 
24  x  12000000)  5  4) 

=  0.0178125  foot  =  0.21375  inch; 


D  = 


and  the  total  deflection  =0.21375  +  0.17086  =  0.38461  inch. 
To  find  the  deflection  at  the  third  loaded  point  due  the  5 
equal  weights  in  their  positions,  we  use  (203),  taking  r  =  3  ; 
thus, 


=  0.0104625  foot  =  0.12555  i 

And  the  deflection  at  this  third  interval,  due  the  beam's  own 
weight  of  27  pounds  per  linear  foot,  is,  by  equation  (190),  put- 
ting x  =  9,  w  —  27, 

27 

y  =  —  -  (6  X   is2  X  o2  —  4  X   15   X  o3  +  O4) 

24  x   i  2000000  v 

=  0.006766  foot  =  0.081192  inch. 
Therefore  the  total  deflection  at  the  third  interval  is 
0.125550  -f-  0.081192  =  0.206742  inch. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  169 


SECTION  2. 

Deflection  of  a  Beam  of  Uniform  Cross-Section,  supported  at  its  Free 

Unfixed  Ends. 

72.  Deflection  due  the  Beam's  own  Weight,  supposed 
to  be  Uniform.  —  For  the  cases  in  this  section  we  employ  the 
same  notation  as  that  given  in  article  17,  Fig.  9,  excepting  that 
we  take  the  origin  of  co-ordinates  at  Olt  a  point  in  the  neutral 
surface,  instead  of  using  O  as  before,  in  order  that  y  may  be 
the  deflection  of  the  neutral  line,  as  it  is  in  the  expression  for 

dy 

the  moment    of   the  internal   forces,  R  —   —El-r-;  =  M,  of 

dx* 

article  63.     We  now  have  x  positive  to  the  right,  and  y  posi- 
tive downwards. 

From  equations  (49)  and  (187), 


dy 
Integrating,  with  the  condition  that  -j-  =  o  when  x  = 


Integrating  again,  with  the  condition  that  jj/  —  o  when  x  =  o, 
we  have,  after  reducing, 

(2°5) 


which  is  the  deflection  of  the  uniformly  loaded  beam  at  any 
point,  w  being  the  load  per  unit  of  the  beam's  length,  /. 


MECHANICS  OF   THE   GIRDER. 


If  in  (205)  we  put  x  —  \l,  we  have 

(206) 


the  deflection  at  the  centre  due  a  continuous  uniform  load, 
Iw. 

EXAMPLE.  —  Beam  of  oak,  54  pounds  per  cubic  foot. 
Length  15  feet  =  /.  Rectangular  cross-section,  depth  i  foot 
=  h  ;  breadth  \  foot  =  b.  E  —  2,000,000  pounds  per  inch. 

.'.    E  =  288000000  pounds  per  square  foot, 
/  =  -fabh*  =  ^,     El  =  1  2000000; 


all  dimensions  in  feet. 

Weight  per  foot  of  length  =  i  X  J  X  i  X  54  =  27  pounds. 

Deflection  due  beam's  own  weight  at  a  point  5  feet  from 
either  end,  by  (205),  is 

y  =  -  2J-  -  (54  -  2  X  15  X  53  +  i53  X  5) 
24  X   12000000 

=  0.001289  foot. 
From  (206),  the  central  deflection  is 

=    5  X  27  X  15*    =  foot_ 

384  x   12000000 

73.  Deflection  due  a  Concentrated  Load,  Wt  placed  at 
the  Horizontal  Distance  a'  from  the  Origin  or  End  of  the 
Beam.  —  When  x  <  a',  equations  (40)  and  (187)  apply;  that  is, 


dx* 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND. 


Let  a  be  the  angle  of  inclination,  or  slope,  of  the  beam  at 
point  of  a 
when  x  =  a', 


the  point  of  application  of  W.     Then,  integrating,  -r  =  tan  a 


Again,  y  =  o  when  x  =  o, 


But  when  #  >  «',  use  equations  (43)  and  (187),  giving 


dy 
Integrating  between  the  limits  tan  a  and  -j-,  and  a'  and 


Integrating  again  between  the  limits  o  and  y,  and  /  and  x,  there 
results 

-  (x  -  /)tan«] 


By  putting  ^  =  a  and  j/  =  jx  in  equations  (207)  and  (208), 
we  find 


MECHANICS  OF  THE  GIRDER. 


Putting  this  value  of  tan  a  in  (207),  we  get,  after  reducing, 
y  =  W(*~\(2l  -  W*  -  *»],  (209) 


which  is  the  deflection  at  any  point  between  the  origin  and  the 
weight  W. 

If  x  =  a',  we  have  the  deflection  at  the  loaded  point, 


And  if  x  =  a'  =  \l, 

tan  «  =  o  ; 
and 


Deflection  at  centre  =  D  =  -~,  (211) 


which  is  a  maximum,  since  £F  is  now  at  the  centre. 

Comparing  (211)  and  (206),  where  wl  =  W  =  entire  load 
on  the  beam,  we  see  that  the  deflection  at  the  centre  due  the 
load,  Iw,  uniformly  distributed  continuously,  is  five-eighths  of 
the  deflection  at  the  centre  due  the  same  amount  of  load  con- 
centrated at  that  point. 

By  putting  /  —  d  for  a'y  and  I  —  x  for  xt  in  (209),  or  by 

substituting  the  value  of   tana  =  "Wj*{  --  a'  I  +  ftf'2J  in 
(208),  we  have 

/2~'z'z)(/~*)~(/~*)3]'    (2I2) 


which   is   the  deflection  when  x  >  a'  \  that  is,  at  any  point 
between  the  weight  W  and  the  right-hand  support. 
I 

EXAMPLE.  —  Take  a  beam  of  pine  weighing  40  pounds  per 
cubic  foot,  of  rectangular  cross-section.  Depth  —  h  =  i8j 
inches,  breadth  =  b  =  15  inches,  length  =  12  J  feet  =  150 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  173 

inches.      Call  E  =   1,680,000  pounds  per  square  inch,  I  = 

-i2^3  =  '   ~1T~  ~  =  7>9l4.$3i2$,  El  —  168  X  79,145,312.5  ; 

f  ,  18.5  X  15  X  40 

beam  s  own  weight  per  inch  of  length  =  w  =  -  —3  - 

=  6.4236^-  pounds.      Deflection  due  beam's  own  weight,  Iw, 
at  a  point  48  inches  from  one  end  is,  by  (205), 

-  6.4236$  -  (484-2X  150x48'  +150*  X  48) 
24  X  i68x  79145312.5 

=  0.0027  incn- 

Deflection  at  centre,  from  beam's  weight,  by  (206),  is 

5  x  6.4236^  x  150* 
D   -  384  X  168  X  79i4S3".S  =  °-°°32  mCh' 

which  is  a  maximum. 

Deflection  at  the  point  of  application,  due  weight  W  = 
17,935  pounds  placed  ar  —  48  inches  from  end  of  beam,  is,  by 
(210), 


D  =  -    o  -  X  (150  -  48)*  =  0.07185  inch. 

3  X   168  X   79145312.5 

Deflection  at  the  centre  when  W^  =  17,935  pounds  is  placed 
48  inches  from  one  end,  is  found  by  equation  (212),  making  x 


17935  X  48  /  IS°2\ 

v  =  -,  1  iqo2  —  482 =  0.078017  i 

12  x  168  X  79i453i2.5\  3  4    / 

And  when  W  is  at  the  centre,  the  central  deflection  is,  from 
(211), 

D  —   -r J-^~ — =  0.0948  inch. 

48  X   168  X   79145312.5 

Add  deflection  due   beam's   own  weight   for  total   maximum 
deflection  =  0.098  inch. 


1/4  MECHANICS  OF  THE   GIRDER. 

74.  Deflection  due  a  Partial  Load,  ix/bt  Uniformly  Dis- 
tributed Continuously  over  the  Length,  b,  beginning  at  the 
Horizontal  Distance,,  #,  from  the  Origin,  Olt  or  Left  End 
of  the  Beam,  Fig.  9. 

To  find  this  deflection,  we  use,  when  x  <  a,  equations  (53) 
and  (187),  giving 

d*y  /  _  (a  +  U) 

J£jg  =  -w'b-    ^-f-    -4  =  -«*  (say). 

Let  a  be  the  angle  of  inclination,  or  slope,  of  the  beam  at  the 
distance  a  from  the  left  end  ;  then  integrating,  with  the  condi- 

tion that  -£•  =  tan  a  when  x  =  a, 
dx 


dx 
Again,  y  =  o  when  x  =  o, 

.-.     EI(y  -  x  tana)  =  -*J°?-  -  a*x\          (213) 

2\3  / 

Let  y  •=.  yT  when  x  =.  a, 

-a-^).    (214) 


But  when  x  >  a  and  <  (a  +  6),  equations  (55)  and  (187)  are  to 
be  employed,  yielding 


DEFLECTION,   END  MOMENTS,   ETC,,   FOUND.  175 

And,  if  J3  is  the  angle  of  inclination  at  the  distance  (a  +  b) 
from  the  origin,  we  may  integrate  as  follows  : 

-^  —  tan  a  when  x  —  a, 


=  gy*»  -  «A  _  L±_£^(«.  _  „.)  +  £^(*  _  a). 

y  —  y^  when  ;r  =  a, 

-     £/|>  -*  -(*-«)  tana]  =  ™^°^-^  -  a*(x  -  a)  | 
_^^_a2(x_a)^^^ 


Let  j  =  y2  when  ^  =  a  +  #,  Fig.  9  ;  so  that,  after  redu- 
cing, (215)  becomes 


t  -y,-b  tan  «)  =          -  +         +       __.      (2l6) 

/     \  2  12  12  2  8  / 

Or,  we  may  integrate  in  a  different  manner  ;  first,  with  the  con- 

dition   -^  =  tan  /?  when  x  =  ^  +  ^, 
dx 


=  g/^3_(a  +  ^3) 

2    (  3  )  2          ( 

Also  j  =  j/2  when  ^r  =  «  +  #, 


i;6  MECHANICS  OF   THE    GIRDER. 

But  in  (217)  y  =  y^  when  x  =  a  ;  therefore,  after  reducing,  we 
have 


EI(y.  -y,+  Jtan/Q  =  +        +      -      -  (2i8) 

/          2  12  6  2  2 


12  2  24 


For  the  remaining  part  of  the  beam,  Fig.  9,  that  is,  when 
x  >  (a  +  £)>  equations  (57)  and  (187)  give 


x  -  a  -  ty)  =  (w'b  -  e)x  -  w'b(a 

—  =  tan  6  when  ^r  =  a  +  <^, 
dx 


-  (a 


zr  o  when  ^r  =  /, 


_(*_/)  tan  )8] 

_  /3 


+  6)  (a  +  tf)  (X  -  /),     (219) 
which  becomes,  if  we  put  y2  for  y,  and  a  +  £  for  ;r,  and  reduce, 

^[>2-  (a  +  l>-  /)tan^8]  =  —(l-a  -  &)*(a  +  ^).     (220) 

From  equations  (214),  (216),  (218),  and  (220),  we  may  now 
determine  the  four  quantities,  tan  a,  tan  /?,  ylt  y2,  so  that  they 
can  be  eliminated  from  (213),  (215),  (217),  and  (219). 


_-- 
\  3        26      24  2636 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  177 


= 

6          8 


3aW    lafrl    n&l  .a*l*  .abl*  .Pl*\       ,       N 
-  ----  1  ---  1  ---  1  —  —  ).      ^224) 
2  6         24        3         2         6  / 

We  have,  then,  from  (213),  where  x  is  not  greater  than  ay 

,     (225) 


which  is  the  deflection  due  w'b  at  any  point  between  the  origin 
and  the  beginning  of  the  partial  continuous  uniform  load  it/b, 
Fig.  9. 

For  the  uniformly  loaded  part,  &,  of  the  beam,  we  find,  from 

(215), 


which  is  the  deflection  due  «/£  at  any  point  of  the  loaded  por- 
tion b,  since  x  is  here  not  less  than  a,  nor  greater  than  a  +  b, 
Fig.  9. 

Equation  (219)  gives  the  deflection  for  the  remaining  part  of 
the  beam,  that  is,  where  x  is  not  less  than  a  +  b  ;  and  we  find 


+  (*-/)  tan  &     (227) 

which  is  the  deflection  due  ii/b  at  any  point  between  the  right- 
hand  end  of  the  beam  and  the  load  w'b,  Fig.  9. 


1 78  MECHANICS  OF   THE   GIRDER. 

If  x  =  a  —  b  —  \l  in  (225)  or  (226),  we  have  the  central 
deflection  when  one-half  of  the  beam  is  uniformly  loaded  con- 
tinuously ;  viz., 

y    =  5*"' 

"    2   X 

which,  if  ze/  =  w,  is  one-half  the  deflection  found  by  (206)  for 
the  fully  loaded  beam. 

The  same  result  may  be  obtained  from  (226)  or  (227)  by 
putting  x  =  b  =  £/,  and  a  =  o ;  for  then  the  one-half  load  is 
upon  the  other  end  of  the  beam.  The  greatest  deflection  due 
a  partial  uniform  load  evidently  occurs  when  the  centre  of  the 
load  and  centre  of  the  beam  are  in  the  same  vertical  line ;  that 
is,  when  a  +  \b  =  J/,  a  =  \(l  —  b\  and  b  =  I  —  2a.  Then, 
putting  x  =  \l  in  (226),  we  may  find  the  greatest  deflection  a 
partial  uniform  load  can  produce. 

But  if  it  is  required  to  find  the  maximum  deflection  of  the 
beam  when  a  given  partial  uniform  continuous  load  has  any 
given  position  upon  it,  we  may  differentiate  (225),  (226),  or 

(227),  put  -^  =  o,  and  so  find  a  value  of  x  that  shallr  ender^  a 

c£Jv 

maximum.  If  we  then  add  the  deflection  at  the  point  so  found, 
due  the  beam's  own  weight,  we  have  the  total  deflection. 

75.  An  important  application  of  (226)  and  (227)  may  be  made 
if  we  take  a  =  o ;  for  in  that  case  the  partial  uniform  continuous 
load  begins  at  the  left  end  of  the  beam,  so  that,  by  assigning 
successively  increasing  values  to  b>  we  may  find  the  deflection 
at  any  point  due  an  advancing  continuous  uniform  load  w'b. 

If  a  =  o, 


jr  =  o,  tana  =  -^-TJ(P  -  tfl  +  4/2), 

and  (226)  becomes 

w'    ( 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  179 

which  is  the  deflection  at  any  point  of  the  loaded  part  of  the 
beam,  where  x  is  not  greater  than  b. 
Also,  if  a  =  o, 


and  (227)  becomes 


which  is  the  deflection  at  any  point  of  the  unloaded  part  of  the 
beam,  where  x  is  not  less  than  b. 

EXAMPLES.  —  Partial  uniform  continuous  load,  it/b.  Wrought- 
iron  15-inch  I-beam.  Length  30  feet  =  360  inches  =  /.  Mo- 
ment of  inertia  /  =  -^(bjij  —  bjif),  by  Table  III.  5. 

Let  h2  =  15-  inches,  b2  =  5|  inches,  h^  =  I2|  inches,  bl  =  4| 
inches,  Fig.  91 ;  putting  /i2  for  ky  and  b2  for  b,  to  avoid  confusion 
here.  Beam  supported  at  ends.  Load  w'  =  75  pounds  per  inch 
of  the  length  b> 

.'.    I  =  TV(5-375  X  153  -  4.75  x  12.753)  =  691. 
Take  E  =  24,000,000,  Table  II., 

/.    El  =  16584000000, 

all  dimensions  to  be  in  inches.     Let  the  load  cover  the  first  10 
feet  of  the  beam. 

ist,  What  is  the  deflection  at  the  end  of  the  load  ? 

We  have  x  =  b  =  -J/  =  120  inches;  and  (228)  applies, 
giving 


75  X  36o4        /i  1511  i  i 

i6584oo 
0.23444  inch. 


«y  — _ I    T          *     vv    \^f    _    \s    

24 X 16584000000^81 


180  MECHANICS   OF   THE   GIRDER. 

2d,  What  is  the  deflection  at  the  centre  of  the  beam  ? 
We  have  from  (229),  if  x  —  J/,  and  b  =  J/, 

75  X  £  X  36o4    (73 
24  x  16584000000!  8  4 

=  0.24421  inch. 

3d,  What  is  the  deflection  10  feet  from  the  unloaded  end  of 
the  beam  ? 

Here  we  use  (229)  also,  putting  x  —  |7,  and  b  =  J/, 

.  75  X  I  X  360*    j  19  5 

24  X   1 65 84000000  (  27  n  9 

=  0.19176  inch. 

4th,  Suppose  it  is  now  required  to  find  the  point  of  greatest 
'deflection  due  this  same  load  of  75  pounds  to  the  inch  on  10 
feet  of  one  end  of  the  beam. 

Differentiating  (229),  we  find,  since  b  =  4/, 


omitting  constants.     Putting  this  value  of  -^  equal  to  zero,  we 

MM? 

at  once  have  x  =  0.43892^  which  is  the  point  of  greatest 
deflection  ;  and,  by  placing  this  value  of  x  in  (229),  there 
results  y  =  0.24847  inch,  which  is  the  greatest  deflection  of 
the  beam  due  this  load  along  one  end. 

5th,  But  if  this  same  load  be  moved  along  to  the  centre,  so 
that  we  have  a  =  \l  =  b,  we  find  the  greatest  deflection  the 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  l8l 

load  can  produce,  by  putting  x  =  \l  in  equation  (226),  where 

yt  becomes  =    IIW   *     and  tan  a  =  /  n    *     from  (223)  and 
1944^7  648^7 

(221).     Thus  (226)  becomes 


75  X  36Q4   )  i  /  i   _I\_,  _!/£  _  £ 


2  x  16584000000(1216 


_I\_,  _!/£  _  £\ 
8i/   8i\2   37 


Deflection  at  centre  =  y  =  0.50063  inch,  \w'l  central ; 
deflection  at  centre  =  y  =  0.24421  inch,  Jo//  at  either  end  ; 

/.    y  =  0.50063  4-  2  x  0.24421  =  0.98905  inch, 

equals  deflection  at  centre  when  the  given  15 -inch  I-beam  of  30 
feet  between  supports  is  loaded  with  13.5  tons,  uniformly  dis- 
tributed continuously.  And  this  result  accords  exactly  with 
that  given  by  (206) ;  thus, 

5  x  75  x  360* 
=  384  X  16584000000  =   °'989°5  mch; 

where,  as  in  other  values  of  the  deflection,  we  have  retained 
several  unnecessary  decimals,  in  order  to  test  the  accuracy  of 
the  solutions. 

6th,  If  this  beam  is  half  loaded  with  75  pounds  to  the  inch, 
we  have  in  (228),  for  the  deflection  at  the  centre,  x  =  b  =  \l 
=  1 80  inches  ;  and 

_         75><36o4        I  i  131/1  i  i\i 

"24X16584000000(16  24      8     \i6  8  4/2 

=  0.494525  inch, 

which  is  half  the  deflection  due  the  fully  loaded  beam,  as  just 
found. 


1  82  MECHANICS  OF   THE   GIRDER. 

7th,  The  maximum  deflection  due  this  half-load  on  one  end 
of  the  beam  is  found,  both  in  position  and  magnitude,  by  differ- 

entiating (228),  putting-^  =  o,  and  solving  the  resulting  cubic 
clx 

equation,  putting  b  =  J/,  /  =  360.     Thus,  omitting  constant 
factors, 


=  4*3  _       x  3<MC.  +         x 
.*.     a:3  —  405^  +  6561000  =  o. 

Solving  this  equation  by  Horner's  Method,  we  find  the  three 

values, 

x  =       165.52  inches, 

x  =      352.08  inches, 
x  =  —112.60  inches. 

But,  since  x  must  be  positive  and  not  greater  than  £/  =  180, 

the  value  here  sought  is 

t 
x  =  165.51995, 

retaining  decimals.  Hence  the  point  of  greatest  deflection  is 
within  the  loaded  part,  and  is  180  —  165.51995  =  14.48005 
inches  from  the  centre  of  the  beam. 

Putting  this  value  of  x  in  (228),  we  find  the  maximum 
deflection  y  =.  0.49855  inch. 

8th,  The  beam's  own  weight  per  inch  of  length,  calling 
wrought-iron  five-eighteenths  pound  to  the  cubic  inch,  is  T5g-  X 
area  of  cross-section  =  -f%(bjit  —  bji^)  =  -^(5.375  X  15  — 
4.75  X  12.75)  =  5-573  pounds,  which,  substituted  for  w  in 
(206),  gives  the  deflection  at  the  centre  due  the  beam's  own 
weight  =  0.07349  mcn  5  s°  that  the  total  central  deflection  for 
the  fully  loaded  beam  is  0.98905  -f  0.07349  =  1.06254  inches. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  183 

76.  To  find  the  Deflection  at  any  Point,  #,  due  any 
Number,  (rx  —  r2),  Equal  Weights,  W,  placed  at  Equal 
Intervals,  c,  along  the  Beam,  the  First  Weight  being  Dis- 
tant by  One  or  More  Entire  Intervals,  c,  from  the  Origin 
or  Left  End  of  the  Beam.  —  For  all  the  weights,  (r  —  r2)  in 
number,  between  the  left  end  and  the  point  x,  we  use  equation 
(212),  which  reduces  to 


y  = 


Now  let  a'  take  the  successive  values  c(r2  -f-  i),  c(r2  -f-  2), 
C(r2  +  3)>  •  •  •  C(r2  +  f  —  r2)>  and  we  have,  by  summing, 


'  =  c(r2  +  i  +  r2  +  2  +  r2  +  3  +  .  .  .  +  r) 


Hence  (212)  becomes 


i)2  -  r22(r2  +  i 

(r  -  r2)(r  +  r2  +  i)(/-  ^)3,      (230) 


which  is  the  deflection  due  r  —  r2  equal  weights,  W,  at  any 
point,  xf  between  the  rth  weight  and  the  right  end  of  the  beam  ; 
r2  being  the  number  of  full  intervals  vacant  at  the  left  end,  and 
x  being  not  less  than  cr. 

For  the  rl  —  r  equal  weights  between  the  point  x  and  the 
right  end  of  the  beam,  we  employ  (209),  which  reduces  to 


1 84  MECHANICS  OF  THE   GIRDER. 

If  in  this  equation  a'  takes  the  successive  values  c(r  +  i), 
c(r  +  2),  c(r  +  3),  c(r  +  4),  .  .  .  c(r  +  r,  -  r),  then,  by  sum- 
ming, we  find 

^ar   =  <-[(r  +  i)  +  (r  +  2)  +  (r  +  3)  +  •  •  •  +  r,] 
=  \c(r,  -  r)(rt  +  r  +  i), 

+  i)2  +  (r  +  2)2  +  (r  +  3)2  +  •  •  •  n2] 
(r,  +  i)(2r,  +  i)  -  r(r  +  i)(2r  +  i)], 

+  i)s  +  (r  +  2)3  +  (r  +  3)3  +  .  .  .  rxa] 
=  i*3[>,a(rs  +  i)2  -  7*(r  +  i)2], 
Sa/0  =  r,  -  r. 

Hence,  for  this  case,  (209)  becomes 
W 


y 

2.^22,11    { 

+  IXr,  -  r)(r«  +  r  +  i)  -  4(r,  -  r)/]^J,      (231) 

which  is  the  deflection  due  the  r,  —  r  equal  weights  on  the 
beam  at  any  point,  x,  between  the  left  end  and  the  (r  +  i)th 
weight ;  x  not  being  greater  than  c(r  +  i). 

Adding  the  deflections  given  by  (230)  and  (231),  and  calling 
their  sum  also  y,  we  have 

W    (r    , 
y  =  rrSwU^fr  -  r«)<r«  +  r'  +  0  -  4(^i  - 


—  6c(r  —  r2)(r  +  r2  + 
i  --  ra)(r,  +  r2  +  i)  +  ^[^(r,  +  i)2  -  r22(r2  +  i)2] 
i)(2rr  +  i)  —  r(r  +  i)(2r  +  i)]^ 

+  i)2  -  ^2(^2  +  i)2]l,    (232) 


DEFLECTION,   END   MOMENTS,  ETC.,   FOUND. 


I85 


which  is  the  deflection  at  any  point,  x,  due  the  r,  —  r2  equal 
weights,  W\  where  r,  denotes  the  number  of  intervals  between 
the  last  weight  and  the  left  end  of  the  beam,  r  a  number  of 
full  intervals  not  less  than  r2,  the  number  of  unloaded  intervals 
at  the  left  end  of  the  beam,  nor  greater  than  TV 

If  in  (232)  we  put  c  =  /  -f-  »,  x  =  -J/,  r2  =  o,  r,  =  ;z  —  I, 
and  r  =  £#  when**  is  even,  but  r  =  J(w  —  i)  when  n  is  odd, 
we  shall  find 


D 


(5«2  -  4),  n  even, 


Wt* 


—  4«2  —  i),  #  odd, 


(233) 


which  is  the  deflection  at  the  centre  due  the  r,  =  «  —  i  equal 
weights,  H7,  covering  the  beam  of  ^  equal  intervals  (/  -f-  «). 

EXAMPLES.  —  Let  us  take  the  same  1 5-inch  I-beam  we 
employed  in  the  examples  of  article  75,  for  which  /  =  691, 
E  =  24,000,000,  /  =  360  inches.  Take  3  weights  of  4,500 
pounds  each,  placed  at  intervals  of  60  inches,  beginning  at  one 
end  of  the  beam ;  then  the  deflection  at  the  centre  is  given  by 
(230)  if  we  put  W  —  4,500,  /  =  360,  c  —  \l  —  60,  r2  =  o, 
r  —  3,  x  =  |/,  El  —  16,584,000,000.  Thus, 


= 


4°°  X 


36o3__/ 


24  X   16584000000^         6 


r*      \/  XX        "»       N/        >l        \/ 

2A-X3X4X — 


—  2X- 
26 


X3X4X 


-  j  =  0.6154  inch. 
8/ 


If  2  more  equal  weights  are  added  at  the  same  interval, 
so  as  to  cover  the  beam,  the  central  deflection  due  these  last  2 
is,  by  (231),  where  r,  =  5,  r  =  3, 


1 86  MECHANICS  OF   THE    GIRDER. 

4=500  X    36o3        (if  \i 

-  24  X   i6584ooooool4  X  6(25  -  9  +  5  -  3)i 

-  2  X  -If  5  X  6  X  ii  -  3  X  4   X   7\-  +  r  I  25  x  36  -  9  x  i6V 
36\  /2       63\  /2 

+  2  X  1(25  -  9  +  5  -  3V  -4  X  2  x  *  j  =  0.3517  inch. 

If  we  compute  the  central  deflection  due  these  5  equal 
weights  by  (233),  we  have  n  —  6,  and 

_         45°°  X  3603       /5  x  36  -  4\  _ 
=  384  x  16584000000^  6       ~ /       °'967i      cn> 

which  is  the  sum  of  the  deflections  found  by  (230)  and  (231). 

Again,  if  there  are  8  weights  upon  the  beam,  each  equal  to 
W  .=  3,000  pounds,  at  intervals  of  40  inches,  we  have  n  =  9, 
/  =  360  inches ;  and  (233)  gives  the  central  deflection, 

3000  x  3602       /5  x  94  -  4  x  92  -  A 
D   =  384  X  16584000000^-          -JT-          -)  =  °'97926  mch. 

In  these  examples  the  weight  has  been  purposely  chosen 
equal  to  75  pounds  to  the  inch  for  the  entire  length  of  the 
beam,  except  a  half-interval,  (/  -f-  2w),  at  each  end ;  so  that 
we  may  compare  the  results  with  the  central  deflection  of  the 
same  beam,  computed  by  (206)  for  the  continuous  uniform 
load  of  75  pounds  to  the  inch,  which  deflection  we  have  found 
to  be  0.98905  inch. 

Now  it  will  be  found  that  the  central  deflection  due  the  dis- 
continuous load,  (n  —  i)  W,  at  equal  intervals,  (/  -=-  n),  will  be 
less  than  that  due  the  continuous  uniform  load,  /w,  until  n 

becomes  infinite,  and  W=  —  infinitesimal,  when  (233)  becomes 

n 

identical  with  (206). 


DEFLECTION,   END   MOMENTS,   ETC.,   POUND.  l8/ 

The  greatest  difference  between  the  central  deflections  of 
these  two  loads,  (;/  —  i)  W  and  Iw,  manifestly  occurs  when 
n  =  2  ;  that  is,  when  there  is  but  one  weight,  and  that  at  the 

centre,  and  equal  to  W  =  —  .      Equation  (233)  then  becomes 

D  =  -4—  —  ,  which  is  four-fifths  of  the  deflection  due  Iw  con- 
384^7 

tinuously  distributed  uniformly,  as  shown  by  (206). 

From  these  considerations  it  appears,  that,  in  practice,  the 

Nformulse  found  in  article  75,  for  a  uniform  continuous  load,  are 

applicable  to  a  uniform  load  distributed,  as  above,  discontinu- 

ously,  or  by  panel  weights,  each  equal  to  (Iw  -f-  «)»  provided  n 

is  large. 

But  in  any  case,  whether  there  be  many  or  few  intervals,  we 
may  find,  by  means  of  equation  (232),  the  greatest  deflection 
due  any  partial  or  complete  load  of  equal  panel  weights,  Wt  and 
the  point  where  it  occurs. 

For  this  purpose,  differentiate  (232)  with  respect  to  x,  and 

put  -f  =  o.     This  gives 
ax 

[6f(rt  -  ra)(rx  +  r2  +  i)  -  12  (r,  -  r)/>2 

—  i2<r(r  —  r2)(r  +  r2  +  i)& 

+  4/V(r,  -  r2)(rt  +  r2  +  i)  +  ^[^(r,  +  i)2  -  r22(r2  +  i)2] 
-  2&2[rl(rl  +  i)(2rx  -f  i)  -  r(r  +  i)(2r  +  i)]  =  o,     (234) 

from  which  we  find 

x  =  A  ±  \/A2  +  B,  (235) 

where 


~  c(r*  -  ra)(rx  -f  r2 
and 


1 88  MECHANICS  OF   THE   GIRDER. 

Now  put  cr  for  x  in  (234) ;  and  find  r  by  trial,  easily,  since  it  is 
an  integer,  and  the  point  of  greatest  deflection  is  approximately 
known,  by  inspection,  for  any  given  load.  Then,  having  r, 
compute  x  in  (235),  after  which  the  greatest  deflection,  j,  may 
be  found  by  equation  (232). 

EXAMPLE  i.  —  Given  the  wrought-iron  I-beam  of  article  75, 
where  I  =  691,  E  —  24,000,000,  /  =  360  inches ;  and  let  there 
be  upon  it  5  weights  of  4,500  pounds  each,  at  equal  intervals  of 
c  =  \l  =  60  inches.     We  then  have  W  —  4,500,  rt  =  5,  r2  = 
o ;  and,  by  putting  cr  for  x  in  equation  (234),  we  find 

2r3  —  i8r2  +  r  +  105  =  o. 

By  trial,  we  see  that  r  =  3,  as  we  should  also  infer  from  the 
symmetrical  load.  Making  r  =  3  in  (235),  there  results  x  =  \l. 
This  value  of  x  placed  in  (232)  gives  the  deflection  y  =  0.9671 
inch,  as  by  (233). 

EXAMPLE  2.  —  If  on  this  same  beam  we  have  4,500  pounds 
at  the  end  of  the  second  and  third  intervals,  we  have  W  = 
4,500,  rI  =  3,  r2  =  i,  c  =  \l  =  60  inches ;  and,  by  putting  cr 
for  x  in  equation  (234),  we  find 

6/3  —  48r2  +  $gr  +  143  =  o, 

where,  by  trial,  we  see  that  r  lies  between  2  and  3.  Making 
r  =  2  in  (235),  we  find  x  =  0.481 41 /.  With  this  value  of  x, 
(232)  gives  the  maximum  deflection  due  the  2  given  weights, 
y  =  0.48934  inch. 

EXAMPLE  3.  —  If  these  2  equal  weights  are  at  the  end  of  the 
third  and  fourth  intervals,  then  r,  =  4,  r2  —  2,  c  =  60 ;  and  we 
shall  find  r  =  3,  x  =. 0.518597,  and  the  maximum  deflection,  as 
before,  y  =  0.48934  inch. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  189 


SECTION    3. 

The  Influence  of  Fixed  Ends  upon  the  Deflection  of  a  Beam  of  Uniform 
Cross-Section,  Supported  at  its  Two  Extremities,  which  are  Assumed 
to  be  Level,  and  One  or  Both  of  Them  Fixed  Horizontally  or  Other- 
wise. Determination  of  the  End  Moments  and  Points  of  Contrary 
Flexure.  ' 

77.  The  influence  of  the  end  couples  upon  the  moments  due 
the  other  forces  has  already  been  found,  by  equation  (93),  to  be 

,-•  M2       —      Mr  n/r 

Mx  =  —  ^—  -  l-x  -f-  Mlf 

where  Mt  is  the  left  end  moment,  and  M2  the  right  end  moment, 
of  the  fixed  beam,  Fig.  12. 

Wherefore,  to  find  the  deflection  due   these  end  couples, 
(187)  becomes 


giving  the  first  member  the  positive  sign,  since  Mt  and  M2  are 
here  assumed  to  have  a  tendency  to  deflect  the  beam  upward, 
and  are  negative  relatively  to  the  moments  tending  to  deflect  it 
downwards. 

If  a  =  the  slope  of  the  beam  at  the  centre,  then  -^  =  tan  a 

ax 

when  x  =  £/,  and  the  first  integration  yields 


dx 
Again,  since  y  =  o  when  x  =  o, 

.-.    EI(y  -  *tan.)  =  *"'  ~  MjX- 


I  go  MECHANICS  OF  THE   GIRDER. 

Also  y  =  o  when  x  =  /, 


24^7 

Therefore 


'    (236) 


which  is  the  deflection  due  the  end  moments  in  terms  of  these 
unknown  end  moments.  Now,  since  (236)  has  been  found 
without  assuming  the  ends  of  the  beam  tangent  to  the  line 
drawn  through  the  two  points  of  support,  we  may  suppose  MT 
or  M2  to  vanish,  or  to  be  equal  to  each  other. 

If  M-  =  °-        ^  Hh(x  ~  7}  (237) 

If  M2  =  o,  y  =  ^(^  -  y?  +  2&).  (238) 

If  Mt  =  M,  =  M,    y  =  ^-(lx  -  X*).  (239) 


In  order  to  determine  the  end  moments  in  particular  cases, 
we  must  consider  the  particular  mode  of  loading. 

78.  Load  Continuous  and  Uniform  throughout,  =  w  per 
Unit  of  Length,  /.  —  If  we  add  equations  (236)  and  (205),  call- 
ing the  result  y,  we  have 


-  lx)\ 


(240) 


which  is  the  deflection  at  any  point  of  the  uniformly  loaded 
beam  of  fixed  ends. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  IQI 

If  the  ends  are  both  fixed  horizontally  (that  is,  if  the  tangent 
to  the  curve  is  horizontal  at  each  end  of  the  level  beam),  we 
must  have  J/x  =  M2,  since  the  load  is  uniform.  And,  differen- 
tiating (240), 


-   4/2)    - 


=  o, 


-f 
dx 


dv 

But,  now,  —  =  o  when  x  =  o, 
dx 


Putting  this  value  of  the  end  moments  into  (240),  there  results 


which  is  the  deflection  at  any  point,  #,  of  a  beam  with  ends  fixed 
horizontally,  under  a  continuous  uniform  load,  w,  per  unit  of 
length,  /. 

If  x  =  J/,  we  have  the  central  deflection 


which  is  one-fifth  that  due  the  same  load  on  the  same  beam  with 
its  ends  not  fixed,  as  given  by  (206). 

Since  Ml  =  M2,  the  total  moment  due  Iw  at  any  point,  is,  by 
(49)  and  (93), 


Mx  = 


1  92  MECHANICS  OF  THE   GIRDER. 

And,  if  we  put  this  moment  Mx  =  o,  we  shall  have  x  represent- 
ing the  distance  from  the  left  end  of  the  beam  to  the  points  of 
contrary  flexure,  as  those  points  are  called  where  the  curvature 
changes  from  convex  to  concave  upward. 
Therefore 

x2  -  lx  +  £/2  =  o, 

x  —  0.21  133/    or    0.788677.  (244) 

79.  But  if  the  right  end  of  the  beam  is  fixed  horizontally, 
while  the  left  end  is  not  fixed  at  all,  we  have  Ml  =  o,  and  (240) 
becomes 


y  =       w(x*  "  2/*3  +  /3*}  ~  4M         ;  (245) 

and 


^       24.57  ( 

equal  to  o  when  x  =  /. 

...    ^2  =  ->/2.  (246) 

Hence,  from  (245), 

y  =         (2^  ~  3/^3  +  /3^)j  (247) 


which  is  the  deflection  at  any  point  of  a  beam  horizontally  fixed 
at  one  end,  and  simply  supported  at  the  other,  under  a  uniform 
load  w  per  unit  of  length,  /;  x  to  be  measured  from  the  unfixed 
end. 

Since,  now,  Mt  =  o,  the  total  moment  due  Iw  at  any  point 
is,  from  (49)  and  (93), 

Mx  =  \w(l  -  *)* 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  1 93 

If  Mx  =  O, 

x  =  &  (248) 

which  is  the  distance  of  the  point  of  contrary  flexure  from  the 
free  end  of  the  beam,  under  the  load  Iw  uniformly  distributed 
continuously. 

EXAMPLES. — Suppose  the  wrought-iron  15 -inch  I-beam  3cr 
feet  in  length,  of  the  examples  in  article  75,  to  be  fixed  hori- 
zontally at  both  ends,  and  loaded  uniformly  with  75  pounds  to 
each  inch  of  its  length ;  what  is  the  deflection  10  feet  from 
either  end  ?  We  now  have  /  =  691,  E  =  24,000,000,  /  =  360,, 
x  —  \l  or  |/.  Hence  (242)  gives  the  deflection 

y  =  75   X  360*  X  4 =          6    inch. 

24  X   16584000000  X  8 I 

At  the  centre,  where  x  =  |-/,  (242)  gives 

=  75  X  360*  x  i =  gl  inch^ 

24  X   16584000000  X   16 

which  is  one-fifth  of  that  given  by  (206)  for  beam  with  free 
ends. 

If  only  one  end  of  the  beam  is  fixed,  (247)  gives, 

When*  =  \l,  y= 7$  X  36o4  X  10 =O.39O74  inch. 

24  X  16584000000  X  8 i 

*  =  i/,  75X3604X1  =0.39563  inch. 

24  X  16584000000  X  8 

*  =  f/,  y= 75  X  360*  X  7 =  o.27352inch. 

24  X  16584000000  X  8 i 

x=  151.7846  inches, ^=0.41 14 1  inch,  a  maximum. 


IQ4  MECHANICS   OF   THE    GIRDER. 

80.  Deflection  of  a  Beam  fixed  Horizontally  at  Both 
Ends,  due  to  a  Concentrated  Load,  W,  placed  at  the  Hori- 
zontal Distance  a  from  the  Left  End  of  the  Beam.  —  From 
equations  (40),  (187),  and  (93),  we  have  the  total  moment  due 
W  when  x  is  not  greater  than  a'y 


Mx  =  -*/        =   W-*  -     ^L  +  Ml}    (a49) 


Integrating,  as  in  article  73,  -^  =  tan  a  when  x  —  a', 


-^)  -*('-').  (.50) 


Again,  ^  =  o  when  ^r  =  o, 

/.     ^"/(^  —  x  tan  a) 

#V  _  /)  +  M,  -  MJx*         ,2  \        ^/jc2          ,  \     ,       , 
=  —  i  -  '—L  -  \  -  !(__  -  <Bf*x\  -  MA—  —  a'x\    (251) 

2/  \3  /  \2  / 

But  when  x  is  not  less  than  a',  use  (43)  with  (93)  and  (187), 
giving 


Mx  =  -*/        =          (/  -  .)  -      L,  +  ^,    (252) 


=  -(Waf  +  Mi  -  M2)x  -  (Wo! 

'-2  =  tan  a  when  x  =  a', 
dx 


! 


-(Wa'  +  Ml)(x  -  a').  (253) 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  195 

=  o  when  x  —  /, 
.-.     EI\_y  —  (x  —  /)tan«] 

Wa>  +  llf.  -  MAxt 


_  ,        _ 


I       3 

^^(*--/)     (254) 


Now  7  in  (251)  is  equal  to  j/  in  (254)  when  x  =  #';  there 
fore,  from  (251)  and  (254),  we  find 

—  [WV(f^  +  |/2  -  aft) 
Ell 

+  Jlf,(il>  +  &•  -  a'l)  -  Jf,W  -  |/0]. 


But  in  (250)  we  now  have  —  =  o  when  x  =  o, 

dx 

...     tana  =  —|  Wa'*?-^-1  +  M/—  -  a'l)  -  M2—\.      (256) 

-£Y/  (  2  \  2  /  2    ) 

Also,  in  (253)  -j-  —  o  when  #  =  /, 

/.     tana  =  -±\_W(W*  +  J«7>  -  *V) 
j^y/ 

+  tK^fx  -  J/2)  (^  -  /2)  +  MJ(l  -  a')].     (257) 
From  (255),  (256),  and  (257),  we  find 

M*  =  -^(l-a'Ya',  (258) 

M2  =  -^(l-a')a'*,  (259) 

which  are  the  end  moments  developed  by  the  weight  W  in  any 
position,  a'. 


196  MECHANICS  OF  THE   GIRDER. 

If  the  weight  W  is  at  the  centre,  a'  —  J/,  and 

M,  =  M2  =  -\WL  (260) 

Eliminating  J/n  M2,  and  tana  from  equation  (251),  we  find, 
x  not  being  greater  than  a', 


(261) 

which  is  the  deflection  at  any  point  between  the  weight  W  and 
the  left  end  of  the  fixed  beam  with  ends  horizontal. 

Again,  eliminating  MI}  M2)  and  tan  <x  from  (254),  we  find,  x 
not  being  less  than  af, 


y  = 

+    3#'2A*   -    tf'3/3],        (262) 

which  is  the  deflection  due  W  at  any  point  between  W  and  the 
right-hand  end  of  the  fixed  beam. 

If  x  =  d  =  J/,  both  (261)  and  (262)  reduce  to 


(263) 


--  ^rr> 
192^7 


which  is  the  central  deflection  when  the  weight  W  is  at  the 
centre  of  the  fixed  beam,  and  is  one-fourth  of  that  due  the  same 
load  on  the  same  beam  with  its  ends  not  fixed,  as  seen  by 
equation  (211). 

To  find  one  point  of  contrary  flexure,  we  put  Mx  =  o  in 
equation  (249),  and,  after  eliminating  M,  and  M2>  have 


(264) 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  197 

If  a'  =  V, 

*  =  I/.  (265) 

For  the  other  point  of  contrary  flexure,  put  Mx  =  o  in 
(252),  and  the  result  is 

/(2/  -  a') 

*  =  y^y  •  (266) 

If  a'  =  \l, 

x  =  f/.  (267) 

From  (265)  and  (267)  it  appears  that  when  the  concentrated 
load,  Wy  is  at  the  centre  of  the  fixed  beam,  the  points  of  con- 
trary flexure  are  each  midway  between  the  centre  and  end  of 
the  beam. 

81.  If  the  beam  is  fixed  horizontally  at  the  right-hand  end, 
but  only  supported  at  the  left  end,  we  have  Mt  =  o  ;  while  M2 
may  be  found  from  (255)  and  (257),  since  the  condition  that 

=  o  when*  =  o,  on  which  (256)  depends,  does  not  now  exist. 


.      M  = 


=  —2(a'2  -  I2)af.  (268) 


This  value  of  M2  placed  in  either  (255)  or  (257),  while  Ml  =  o, 
gives 


tana  =  -(^3/a  _  ^'2/3  4.  i<//4  _  ±a's)9        (269) 
EH* 

which  is  the  tangent  of  the  angle  of  inclination  of  the  beam  at 
any  point  where  the  load  W  may  be,  while  only  the  right  end 
is  fixed  ;  af  to  be  measured  from  the  free  end.  With  these 
values  of  Mlf  M2)  and  tana  substituted  in  equation  (251),  we 
find 

W 


12^7/3' 


198  MECHANICS  OF  THE   GIRDER. 

which  is  the  deflection  at  any  point  between  the  weight  W  and 
the  unfixed  end  of  the  beam,  froni  which  end  a'  and  x  are  to  be 
measured. 

In  the  same  manner,  from  equation  (254)  we  find,  x  being 
not  less  than  a', 


y  = 

+  (3*'3t2  +  3^4)*  -  20''3/3],     (271) 

which  is  the  deflection  at  any  point  between  the  weight  W  and 
the  horizontally  fixed  end  of  the  beam  ;  a'  and  x  being  measured 
from  the  free  end. 

If  in  either  (270)  or  (271)  we  put  x  =.  a'  —  J/,  we  have  the 

central  deflection 

n         7  #73 

=  (272) 


due  the  concentrated  load  W  applied  at  the  centre  of  the  beam 
horizontally  fixed  at  one  end. 

If  we  differentiate  (270),  and  put  -^  =  o,  we  shall  find 

*  =  **'"'- 1  $lt£)*          (273) 


which  is  the  point  of  maximum  deflection  between  the  weight 

the  free  end. 
If  the  weight  is  at  the  centre,  a'  =  J/,  and 


x  =  ±/V     =  0.4472I/.  (274) 

In  a  similar  manner,  differentiating  equation  (271),  and  put- 


ting -=£  =  o,  we  find 
dx 


73  +  a'2/ 

2  _    >S  (275) 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  199 

the  point  of  maximum  deflection  between  W  and  fixed  end, 
where  x  cannot  be  less  than  a'  ;  that  is,  a'  in  this  formula 
cannot  be  greater  than  x. 

Putting  at-  for  x  in  (275),  we  may  find  easily,  by  trial,  that 
ci  —  0.414213/13  the  greatest  value  a'  can  have  in  this  case  of 
a  maximum  value  of  y  between  the  weight  W  and  the  horizon- 
tally fixed  end  of  the  beam. 

The  point  of  contrary  flexure  may  be  found  from  (252)  by 
putting  Mx  =  o,  M!  =  o,  and  M2  as  in  (268).  This  substitu- 
tion gives 


If  the  weight  Wis  at  the  centre,  a'  =  ^/,  and 

x  =  T8T/,  (277) 

which  is  the  distance  of  the  point  of  contrary  flexure  from  the 
free  end  of  the  beam. 

If  a?  =z  o,  x  =  |/;  and  if  #'  =  /,.#=:/:  which  are  the 
limits  to  the  range  of  the  point  of  contrary  flexure,  for  a  con- 
centrated load  W,  on  a  beam  fixed  horizontally  at  one  end,  and 
free  at  the  other  ;  x  being  measured  from  the  free  end. 

EXAMPLES.  —  Take  the  1  5-inch  I-beam  of  article  75,  and 
suppose  it  bears  a  concentrated  load  W  =  27,000  pounds, 
and  that  both  ends  are  fixed  horizontally.  We  have,  as  before, 
/  =  691,  E  =  24,000,000,  /  =  360  inches. 

When  W  is  at  the  centre,  what  is  the  deflection  halfway 
between  the  centre  and  either  end  of  the  beam  ? 

Put  a'  =  l/  and  x  =  \l  in  (261),  or  a'  =  \l  and  x  =  \l  in 
(262),  and  find 


=  27000    X  r/a    __          _    2N    1 

6  x  i  65  84000000  LV 


+  (t  -  i  +  f)A]  =  o-i978i  inch. 


2OO  MECHANICS  OF   THE   GIRDER. 

At  the  centre  the  deflection  is,  by  (263), 

D  =        *7°ooX36o3        =  6 

192   X   16584000000 

which  is  one-fourth  of  1.58248  =  the  deflection  due  the  same 
load  on  the  same  beam  with  free  ends.  And  this  1.58248  is, 
again,  eight-fifths  of  0.98905,  the  deflection  found  by  (206)  for 
the  same  load  continuously  distributed  uniformly  over  the  same 
beam  with  free  ends. 

The  points  of  contrary  flexure  are  given,  by  (265)  and  (267), 
at  90  inches  and  270  inches  from  either  end.  Now,  since  the 
deflection  at  the  quarter-points  is  just  one-half  that  at  the  cen- 
tre, it  follows  that,  in  this  case,  the  end  of  the  neutral  line,  the 
point  of  contrary  flexure,  and  the  centre  are  in  the  same  straight 
line. 

When  H^is  at  the  distance  ct  —  |/from  the  left  end  of  the 
beam,  what  is  the  maximum  deflection  ? 

Differentiating  (262),  and  putting   ?  =  o,  we  find 

x  =  o.4/, 
/.    y  =  0.2136  inch. 

Or,  if  af  =  |/,  we  find  in  the  same  way,  from  (261), 

x  =  o.6/, 
/.    y  =  0.2136  inch. 

If  a'  =  |/,  the  points  of  contrary  flexure  are,  by  (264)  and 

(266),  x  =  Ty,  *  =  f  /. 

But  if  a'  =    /, 


Let  us  now  suppose  that  this  beam  is  fixed  horizontally  at 
the  right-hand  end,  but  is  simply  supported  at  the  left  end. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  2OI 

When  W  =  27,000  pounds  is  at  the  centre,  what  is  the 
deflection  at  the  quarter-points  ? 

Putting  a'  =  J/,  and  x  =  J/,  we  find,  from  (270), 

y  =  0.5316  inch. 

But  if  a'  =  J/,  and  ;tr  =  f /,  (271)  gives 
y  =  0.3091  inch. 

H7  remaining  at  the  centre,  the  central  deflection  is,  from 
(272),  D  —  0.69234  inch. 

Also,  from  (274)  and  (270),  the  maximum  deflection  due  W 
at  the  centre  is  y  =  0.70769  inch. 

If  we  place  the  weight  W  =  27,000  pounds  at  the  distance 
a'  =  0.4142 1 3/ from  the  free  end  for  the  maximum  value  of  the 
deflection  y,  we  shall  find,  by  (275),  x  —  a'  =  0.414213/5  and 
from  (270)  or  (271),^  =  0.74534  inch,  which  is  the  greatest 
deflection  W  can  produce  on  this  beam,  since  it  is  at  the  point 
of  maximum  deflection. 

Putting  a'  —  O.4i42i3/in  (276),  we  find 

x  =  0.7071064 

the  point  of  contrary  flexure  when  W  is  at  the  lowest  point 
of  the  beam  fixed  horizontally  at  one  end ;  x  to  be  measured 
from  the  free  end. 

82.  Any  Number,  rl  —  r2,  Equal  Weights,  W,  placed  at 
Equal  Intervals,  c,  along  the  Beam ;  the  First  Weight  being 
(r2  +  i)  Intervals  from  the  Left  End,  and  the  Beam  being 
fixed  Horizontally  at  Both  Ends.  —  Let  r  —  r2  denote  the 
number  of  equal  weights,  and  r  equal  the  number  of  full  in- 
tervals, between  the  point  x  and  the  origin  or  left  end  of  the 
beam,  Fig.  \2  ;  then  rt  —  r  =  the  number  of  weights  between 
the  point  x  and  the  right  end,  if  any. 


2O2  MECHANICS   OF  THE   GIRDER. 

The  deflection  at  the  point  x  due  any  one  of  the  r  —  r2 
equal  weights,  Wy  is  given  by  equation  (262).  Let  a'  in  that 
equation  take  the  successive  values  c(r2  -\-  i),  c(r2  +  2), 
c(r2  +  3),  .  .  .  c(r2  +  r  —  r2)  ;  then,  by  summing,  we  have 

r2[(r2  +  i)2  +  (ra  +  2)2  +  (ra  +  s)2  +   .  .  .   +  r2] 
[r(r  +  i)(2r  +  i)  -  ra(ra  +  i)(2ra  +  i)], 


+  i)3  +  (ra  +  2)3  +  (ra  +  3)2  +...+: 

-  £!r-2(    -i.    v  _     2C-  4-  i  VI 
4 

which  values,  put  in  the  place  of  a?2  and  a'*  in  (262),  give 


(278) 


which  is  the  deflection  due  r  —  r2  equal  weights  at  any  point, 
Xj  between   the  rth  interval   and   the  right   end  of   the  beam 
having  both  ends  horizontally  fixed  ;  x  being  not  less  than  cr. 
If  in  (278)  we  make  x  —  cr,  and  r2  =  o,  then 


+  s)/  -  2c*r*(r  +  i)],      (279) 

which  is  the  deflection,  at  the  rth  weight,  due  r  equal  weights, 
W,  along  the  left  end  of  the  beam  at  equal  intervals,  c. 

Again,  the  deflection  at  the  point  x  due  any  one  of    the 
r,  —  r  equal  weights  beyond  the  point  xy  is  given  by  equation 

(261). 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  2O3 

Let  a'  in  that  equation  take  the  successive  values  c(r  -\-  i), 
c(r  +  2),  c(r  -f-  3),  .  .  .  crl  ;  then  summing  as  in  article  76, 
and  putting  the  values  of  S#'°,  %a',  ^/2,  2<A  into  equation  (261), 
we  find 


'  (r«  +  O^x  Hr.0--,r  t  'X27'  +  J 

_  (ri  _  r)/3  _  ^3[ria(rx  +  i)2  _  (r  4.  i)'r»]  j*s 
02  -  (r  +  i)2r2]/  -  cz[r,(r,  +  i)(2rI  +  i) 
-  (r  +  i)(2r  +  i)r]/2  +  f^(rf  -  r)(rt  +  r  +  i)/*$**l,        (280) 


which  is  the  deflection  due  rr  —  r  equal  weights,  PF,  at  any 
point,  x,  between  the  (r  +  i)th  interval  and  the  left  end  of  the 
beam  ;  x  being  not  greater  than  c(r  -f-  i). 

Adding  equations  (278)  and  (280),  and  calling  the  result  y 
still,  we  have 


i)  -  (ra  +  i)(2ra  +  i 
-  (rx  -  r)/3  -  ^3[rj»(ri  +  x)2  _  (ra  + 
i)2  -  (ra  +  i)2r22]/  -  ^[^(r,  +  i)(2rx 
ra  +  i)(2r2  +  i)r2]/*  +  |r(rx  -  r)  (rx  +  r 
i)(2r  +  i)  -  (ra  +  i)(2ra  +  i)r2]^ 

+  i)2]  -  (ra  +  i)2r22)/3,      (281) 


which  is  the  deflection  due  all  the  r,  —  r2  equal  weights  at  any 
point,  xy  between  the  rth  and  the  (r  +  i)th  intervals  ;  x  being 
not  less  than  cr,  nor  greater  than  c(r  +  i),  while  r  here  is  not 
greater  than  rx,  nor  less  than  ra. 

Beam  fixed  horizontally  at  both  ends.     If  we  now  suppose 

the  beam  divided   into    n  full  intervals,  each  =  c  =  -,  and  a 

n 


204  MECHANICS  OF   THE    GIRDER. 

weight,  Wy  at  each  point  of  division  ;  and  further,  if  we  require 

the  central  deflection  due  such  a  load,  we  have  x  —  J/,  c  =  -, 

n 

rl  =.  n  —  i,  r2  =  o,  r  =.  \n  when  n  is  even,  but  r  =:  %(n  —  i) 
when  n  is  odd. 

Placing  these  values  in  (281),  we  obtain 


n  even, 

(282) 


. 


384^/^3 

which  is  the  deflection  at  the  centre  due  the  r,  =  n  —  i  equal 

weights,  Wy  covering  the  beam  of  n  equal  intervals,  — ;  beam 

n 

fixed  horizontally  at  both  ends. 

The  end  moments,  Mlt  M2,  due  a  single  weight,  W,  are  given 
by  (258)  and  (259),  which  reduce  to 


Now  let  a'  take  the  successive  values  c(rz  +  i),  c(r2  +  2), 
+  3),  .  .  .  c(r2  +  r,  —  r2),  so  that  we  have 


rf  -  ra 


-  r2)(r,  +  r2  -h  i), 

2  4-  i)2  +  (r2  +  2>2  4-  (ra  4-  3)2  +  -  -  -  +  ? 
O'i  +  i)(2rz  +  i)  -  r2(r2  +  i)(2ra  4-  i)], 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  20$ 


+.  !)3  4.  (r2  +  2)s  +  (TZ  +  3)3 
=  M^O,  +  i)2  -  r22(r2  +  i)2], 


OO-,  +  r2  +  i)/2 

i)(2rx  +  i)  -  r2(r2  +  i)(2r2  +  i)]/ 
+  i^O,2^  +  i)2  -  r22(r2  -f  i)2]J,  (283) 


-  Mr,2(rz  +  i)2  -  9«22(r2  +  i)^]J,     (284) 

which  are  the  end  moments  due  rz  —  r2  equal  weights,  W\  both 
ends  of  beams  fixed  horizontally. 

The  greatest  deflection  due  r,  —  r2  equal  weights,  Wy  placed 
at  equal  consecutive  intervals  anywhere  along  the  beam,  may 
be  found  by  the  following  method  :  — 

If  in  equation  (281)  we  provisionally  make  x  =  cr,  we  shall 
have  yr.  Then,  putting  r  +  i  for  r  in  this  value  of  yrt  we  find 
yr  +  !  ;  and  therefore 

Ay  =  >  +  I  -  yr. 


Now,  by  making  by  =  o,  we  obtain  a  value  of  r  the  inte- 
gral part  of  which,  not  less  than  r2  nor  greater  than  rx,  will  be 
the  value  of  r  in  (281)  when  7  is  a  maximum.  Then,  differen- 

tiating (281),  and  putting  -^  =  o,  we  find  a  value  of  x  which 

renders  y  a  maximum. 

Although  this  solution  is  rigorous,  it  need  not  often  be 
employed,  since  (281)  gives  the  deflection  at  as  many  points 
as  we  please,  and  a  close  approximation  to  the  greatest 
value  of  y  may  be  found  by  a  few  trials.  An  example  will 
be  given. 


2O6  MECHANICS  OF   THE   GIRDER. 

For  finding  the  points  of  contrary  flexure,  we  have,  from 
(6  1)  and  (93), 

Mx  =      [(r,  -  r)/  -  \c(r,  -  r)  (r,  +  r  +  i)> 


which   is   the  moment  due  the  rt  —  r  equal  weights   at  any 
point,  x,  between  the  (r  +  i)th  weight  and  the  left  end  of  the 
beam  ;  x  being  not  greater  than  c(r  +  i). 
Also,  from  (60)  and  (93), 

Mx  =  ¥c(r  -  ra)  (r  +  r2  +  i)  (/  -  *)  -  ^  ^  M*x  +  Mt,     (286) 

which  is  the  moment  due  the  r  —  r2  equal  weights  at  any 
point,  x,  between  the  rth  weight  and  the  right  end  of  the  beam  ; 
x  not  being  less  than  cr. 

If  we  now  add  equations  (285)  and  (286),  representing  the 
three  resulting  moments  still  by  the  symbols  MXJ  Miy  M2,  we 
shall  have 


-  r  -  r2)  +  M19       (287) 

which  is  the  moment  due  all  the  r,  —  r2  equal  weights  at  any 
point,  x,  between  the  rth  and  the  (r  +  i)th  weights  ;  x  being  not 
less  than  cr,  nor  greater  than  c(r  +  i),  between  r2  and  rx. 

Now,  at  the  points  of  contrary  flexure,  we  have,  in  (287), 
Mx  =  o ;  and  therefore 

oc  — —  •-- ' 


,  -  r)/  -  ^(rx  -  ra)  (r,  +  r2  +  i)] 

But  in  this  expression  for  x,  whose  value  lies  somewhere 
between  cr  and  c(r  +  i),  we  cannot  tell  what  to  call  r.    Let  us, 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  2O? 

therefore,  in  (288),  put  cr  in  the  place  of  x,  and  determine  the 
values  of  r,  which  must  be  integers ;  we  can  then  find  x,  since 
the  other  quantities  in  (288)  are  given. 
By  putting  cr  for  x  in  (288)  we  obtain 


r  =  -«  ±  y/^:'  -  r,'  -  r,  +  e*,  (289) 

where 

«  =  — l-jj^ — -  —  rt  —  %  +  — -(r,  —  r2)  (r,  +  r2  +  i). 

If  (289)  gives  values  of  r  not  integral,  the  decimals  must  be 
rejected,  and  the  integers  retained.  Equation  (289)  will  give  r 
an  integer  only  when  there  happens  to  be  a  point  of  contrary 
flexure  at  the  rih  interval ;  that  is,  where  x  really  equals  cr. 

Having  r2  equal  intervals,  c,  without  weights  at  the  left  end 
of  the  beam  fixed  horizontally  at  both  ends,  succeeded  by  rl  —  r2 
equal  weights,  we  have  found  the  corresponding  end  moments 
in  equations  (283)  and  (284). 

By  making  r2  =  o  in  those  equations,  there  results 


i)/  -  6/2  -  tfr^rt  +  i)],       (290) 
i)  -  2(2r,  +  i)/],  (291) 


which  are  the  end  moments  due  rx  equal  weights,  W,  placed  at 
equal  intervals,  c,  along  the  beam  fixed  horizontally  at  both 
ends  ;  the  first  weight  being  at  the  distance  c  from  the  left  end. 
83.  Beam  fixed  Horizontally  at  the  Right  End,  and  simply 
supported  at  the  Left,  uniformly  loaded  for  a  Part  or  All 
of  its  Length  with  Equal  Weights,  W,  at  Equal  Intervals, 
c.  —  If  the  first  weight  is  r2  +  i  intervals  from  the  free  end, 
and  if  there  are  r  —  r2  equal  weights,  then  the  deflection  at 


208  MECHANICS  OF  THE    GIRDER. 

any  point,  x,  between  the  rth  interval  and  the  horizontally  fixed 
end  of  the  beam  is  given  by  (271),  provided  we  put  therein 
For  a', 

\c(r  -  ra)(r  +  ra  +  i). 
For  a'\ 

(r  +  i)2-  r22(r2  +  i)2]. 


But  if  the  first  weight  is  at  the  distance  c(r  +  i)  from  the 
free  end,  and  if  there  are  rl  —  r  equal  weights  at  equal  inter- 
vals, c,  beyond,  then  the  deflection  at  any  point,  x,  between  the 
(r  +  i)th  weight  and  the  free  end  of  the  beam  is  given  by 
equation  (270)  if  there  we  substitute 
For  a'°, 

r,  —  r. 
For  a', 

\c(r*  -r)(rs  +  r+  i). 
For  a'2, 

^[rx(rx  +  iX2rx  +  i)  -  r(r  +  i)(2r  +  i)]. 
For  of3, 

(r  +  i)2]. 


If,  then,  we  add  the  two  deflections  thus  derived  from  (271) 
and  (270),  we  shall  have 


+  i)2  -  T22(r2  +  i)2]  -  2(rx  - 

-  3t(r  -  ra)(r  +  r2 
i)2  -  (ra  +  i)2r22]/2  +  f^(rx  -  ra)  (r,  +  r2  +  i)/4 
rx  +  i)(2rx  +  i)  -(r  +  i)(2r  +  i)r]/^}* 

+  !)»  -  (ra  +  i)2r22]/3J,      (292) 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  2OQ 

which  is  the  deflection  due  all  the  rl  —  r2  equal  weights  at  any 
point,  x,  between  the  rth  and  the  (r  -f-  i)th  points  of  division  ; 
x  being  not  less  than  cr  nor  greater  than  c(r  -+-  i),  but  r  from 
r2  to  TV 

In  this  case,  where  Mt  —  o,  M2  is  derived  from  (268),  which 

reduces  to  M2  —  —^  -  a'  I2). 

For  a'  put 

\c(r,  -  r2)(rT  +  r2  +  i). 
For  of*  put 

J*3[>ia(ri  +  i)2  -  r22(r2  +  i)2]. 
We  then  have 


-  Mrx  -  r2)(rI  +  r2  +  i)/«^     (293) 

which  is  the  end  moment  due  r^  —  r2  equal  weights,  W,  uni- 
formly distributed  at  equal  intervals,  c,  on  any  part  of  the  beam 
fixed  horizontally  at  one  end  and  simply  supported  at  the  other  ; 
rl  and  r2  to  be  counted  from  the  free  end. 

84.  Deflection,  End  Moments,  and  Points  of  Contrary 
Flexure,  due  a  Partial  Uniform  Load  continuously  distrib- 
uted, when  Both  Ends  of  the  Beam  are  fixed  Horizontally. 
-  We  might  proceed  in  this  case  as  in  article  74,  using  equa- 
tions (53),  (187),  and  (93);  but,  as  the  process  is  tedious,  we 
employ  the  following  method  instead,  utilizing  results  already 
obtained. 

Let  n  denote,  as  heretofore,  the  whole  number  of  intervals, 
each  equal  to  (/  -f-  n}.     Let  r2  denote  a  certain  part  of  n,  which 

we  will  call  jn  ;  let  rx  =  a  "j"  n,  where  neither  a  nor  a  +  b  can 
exceed  /. 


2IO  MECHANICS  OF   THE   GIRDER. 

Now,  for  a  uniform  continuous  load  we  must  have  in  the 
values  of  MI  and  M2,  equations  (290)  and  (291),  n,  r2,  and  r1 
infinite,  and  W  infinitesimal ;  so  that  we  must  put  n  W  —  w'l 
if  w'  =  the  weight  per  unit  of  the  length. 

Making  these  substitutions  in  (290)  and  (291),  they  become 


-  <*"%     (294) 
(295) 


which  are  the  end  moments  due  the  uniform  continuous  load 
•it/  per  unit  on  the  length  b,  measured  to  the  right  from  a  point 
at  the  distance  a  from  the  left  end  of  the  beam  fixed  horizon- 
tally at  both  extremities. 

Now,  if  a  —  o  (that  is,  if  the  continuous  uniform  load  begins 
at  the  left  end,  and  extends  over  the  length  b),  equations  (294) 
.and  (295)  reduce  to 


(296) 


-  4*0,  (297) 


which  are  the  end  moments  due  the  continuous  uniform  load 
w'  per  unit,  on  the  length  b>  measured  from  the  left  end. 

It  may  be  noted  here,  that  if  in  (296)  and  (297),  while  a  =  o, 
we  suppose  b  =  /,  these  values  of  Ml  and  M2  become  each  equal 
to  —^w'l2,  which  accords  with  equation  (241)  for  the  fully 
loaded  beam. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND. 


211 


Let  us  now  put  the  values  of  Mt  and  M2,  as  given  by  (294) 
and  (295),  into  equation  (236) ;  we  shall  then  have 

wf 


>      (298) 


which  is  that  part  of  the  deflection  due  to  the  influence  of  the 
end  moments,  the  beam  horizontally  fixed  at  both  ends  being 
loaded  with  w'  per  unit  for  any  part,  bt  of  the  beam's  length,  /; 
x  varying  from  o  to  /. 

If  x  be  now  restricted  so  as  not  to  exceed  a,  and  we  add  y 
in  (298)  toy  in  (225),  the  sum  will  be  the  deflection  due  w'b  at 
any  point  between  the  origin  and  the  beginning  of  the  partial 
continuous  uniform  load  w'b. 

If  again  we  limit  x  between  the  values  a  and  a  +  b,  and  add 
the  values  of  y  in  equations  (298)  and  (226),  the  sum  will  be 
the  deflection  due  in  w'b  at  any  point  of  the  loaded  portion  b. 

Finally,  by  making  x  not  less  than  a  +  b  in  (298),  and  add- 
ing that  equation  to  (227),  the  sum  of  the  second  members  will 
be  the  deflection  due  w'b  at  any  point  between  the  right-hand 
end  of  the  beam  and  the  load  w'b. 

It  is  evident,  that,  by  assigning  the  proper  values  to  a  and  b, 
we  may  place  the  load  anywhere  upon  the  beam,  and  give  it  any 
magnitude  not  exceeding  w'l.  Also,  we  may  put  many  partial 
uniform  continuous  loads,  wj>u  wj)^  w^by  etc.,  upon  the  beam, 
by  so  choosing  the  values  of  a,,  a2,  ay  etc.,  bu  b2,  bz,  etc.,  that 
the  partial  loads  shall  take  desired  positions,  whether  they  are 
required  to  be  equal  to  each  other,  or  to  overlap,  or  to  have 
intervals  between  them. 

But  it  is  not  necessary  to  formulate  the  deflection  for  such 
totals  here. 


212  MECHANICS  OF   THE   GIRDER. 

It  remains  to  find  the  points  of  contrary  flexure  for  partial 
continuous  uniform  loads,  ix/b,  when  the  beam  is  fixed  horizon- 
tally at  both  ends. 

If  there  is  a  point  of  contrary  flexure  between  the  left  end 
of  the  beam  and  the  beginning  of  the  partial  load  (that  is,  within 
the  length  a),  we  use  equations  (53)  and  (93),  giving 


n/r           ,jl  —  a  —  *b         Mr  —  M2 
Mx  =  wb *-# -x 


If  Mx  =  O, 


' 


,  -  M2  -  w'b(l  -  a  - 


where  the  values  of  Ml  and  M2  are  to  be  taken  from  (294)  and 
(295),  and  x  cannot  be  greater  than  a.  Should  (299)  yield  a 
value  of  x  either  negative  or  greater  than  a,  there  is  no  point 
of  contrary  flexure  in  the  part  a. 

For  the  loaded  part  of  the  beam  b,  we  have  equations  (55) 
and  (93),  giving 


**•  IT    —  a  — 

Mx  =  wb 


\,  *  —      2          ,,- 

x  —  a}2  --  -  -  -x  +  Ml  =  o, 


.-.     x  =  e  ±  L  -  a*  +  ?,  (300) 


in  _  a  _  U}       M,  —  M2    . 

where  «  =  -± *-«• — 7- — -  +  a. 

/  wl 

M!  and  M2  are  given  by  (294)  and  (295). 

When,  in  (300),  either  value  of  x  is  less  than  a  or  greater 
than  a  +  b,  it  must  be  rejected ;  and  when  both  values  of  x  are 
in  this  condition,  there  is  no  point  of  contrary  flexure  in  the 
loaded  part  b. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  213 

In  finding  the  point  of  contrary  flexure  between  the  right 
end  of  the  beam  and  the  load  w'b,  we  employ  equations  (57)  and 
(93)  ;  taking,  as  before,  the  values  of  Mx  and  M2  from  (294)  and 
(295).  Thus, 

Mx  =  v/b(a  +  \b)1-          -  Ml  ~  Mzx  +  Mt  =  o, 


Mt  -  M2  +  z^O  4-  ty)° 

Equations  (299)  and  (301)  show  that  there  can  be  but  one 
point  of  contrary  flexure  between  either  end  of  the  beam  and 
the  adjacent  end  of  the  load,  while  (300)  indicates  that  there 
may  be  two  such  points  within  the  length  b  covered  by  the 
uniform  load  w'b. 

85.  Partial  or  Full  Continuous  Uniform  Load,  «/£,  on 
any  Portion  of  a  Beam  fixed  Horizontally  at  the  Right 
End,  but  simply  Supported  at  the  Left.  —  Proceeding  as  in 

article  84,  we  make  c  =  -,  r  =  ^n,  rl  =  a    *"    n,  and  substi- 

n          /  / 

tute  these  values  in  (293),  which,  when  n  is  infinite,  and  W 

infinitesimal  and  =  —  ,  becomes 
n 

M*  =  &3(a  +  ^)4  ~  a"  ~  2l^(a  +  b^  ~  ^'     (3°2) 

which  is  the  moment  at  the  fixed  end  due  the  uniform  continu- 
ous load,  wfb,  anywhere  on  the  beam.  Here,  if  a  =  o,  and 
b  —  /,  the  beam  is  fully  covered  by  the  load,  and  M2  =  —  \w'l2, 
in  agreement  with  equation  (246). 

If  in  (302)  a  •=.  o,  we  have  as  the  moment  at  t^e  fixed  end, 
when  the  partial  load  w'b,  begins  at  the  free  end, 

-  (3°3) 


214  MECHANICS  OF  THE   GIRDER. 

Substituting  the  value  of  M2  as  given  by  (302),  in  equation 
(237),  we  obtain 


(304) 


which  is  the  deflection  due  the  end  moment  M2  when  M^  =  o, 
and  the  load  is  w'b  in  any  position  ;  x  varying  from  o  to  /. 

If,  as  in  article  84,  x  be  now  limited  so  as  not  to  exceed  a, 
and  we  add  y  in  (304)  to  y  in  (225),  the  algebraic  sum  will  be 
the  deflection  due  w'b  at  any  point,  x>  between  the  free  end  of 
the  beam  and  the  beginning  of  the  load  w'b. 

If,  again,  x  be  limited  between  the  values  a  and  a  -f-  by  and 
we  add  algebraically  the  values  of  y  in  equations  (304)  and 
(226),  the  result  will  be  the  deflection  due  w'b  at  any  point,  x, 
of  the  loaded  portion  b. 

Also,  by  making  x  not  less  than  a  +  b  in  (304),  and  adding 
that  equation  to  (227),  the  sum  of  the  second  members  will  be 
the  deflection  due  w'b  at  any  point,  x,  between  the  right  or  fixed 
end  of  the  beam  and  the  load  w'b  ;  x  measured,  as  usual,  from 
the  free  end  of  the  beam. 

The  point  of  contrary  flexure  for  the  beam  fixed  horizontally 
at  one  end  and  simply  supported  at  the  other,  which  is  taken 
as  the  origin,  is  found  for  a  partial  continuous  uniform  load, 
w'by  by  means  of  equations  (299),  (300),  and  (301)  for  their 
respective  cases,  by  putting  M,  =  o,  and  taking  M2  from 
(302). 

86.  Examples  illustrating  Articles  82,  83.  —  For  the  sake 
of  comparing  the  deflection  of  the  same  beam  when  one 
or  both  its  ends  are  fixed,  with  its  deflection  when  both  ends 
are  simply  supported,  we  further  consider  the  1  5-inch  rolled 
wrought-iron  I-beam  of  30  feet  clear  span,  whose  moment  of 
inertia  I  =  691,  and  whose  modulus  of  elasticity  E  = 
24,000,000,  as  given  in  article  75. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  21  5 

ist,  Take  3  weights,  of  4,500  pounds  each,  placed  at  inter- 
vals of  60  inches,  beginning  at  the  left  end  of  the  beam  fixed 
horizontally  at  both  ends  ;  then  the  deflection  at  the  centre  is 
given  by  (278)  if  we  put  W  =  4,500  pounds,  /  =  360  inches, 
c  =  i/  =  60,  r  =  3,  and  x  —  \l\  El  being  16,584,000,000. 
Thus, 

4500x36°*     |/IX_LX84_IX_^XI44V 
6  X   16584000000(^2       36  2       216  /8 

+  (3  x  144  _  8_4\i  +  i  x  84  x  i  _  i  x  144)  =  8    inch> 

\4       216       36/4       2       36       2       4       216) 

2d,  If  2  other  equal  weights,  4,500  pounds  each,  be  added 
at  the  same  interval  of  60  inches,  so  as  to  cover  the  beam  with 
concentrated  loads,  the  central  deflection  due  these  last  2  is, 
by  (280),  where  r  =  3,  and  r,  —  5,  or  by  (278),  making  r  =  2, 


inch. 


3d,  For  the  5  equal  weights  now  on  this  beam,  (281)  gives 
the  deflection 


=      4500  x  360*     l/i  x  i  x         _  2  _  2oo\i 
6  X  16584000000(^2        36  432/8 


=  0.19781  inch; 
or,  (282)  gives  the  same  much  more  simply. 

This  value  is,  as  it  should  be,  the  sum  of  the  two  deflections 
last  found. 


2l6  MECHANICS  OF  THE   GIRDER. 

4th,  Suppose  the  fifth  weight  removed  from  the  beam,  what 
is  the  deflection  at  the  fourth  weight  ?  Use  (279),  making 
r  =  4,  c  —  \l  =  60  inches,  W  =  4,500  pounds  ; 

._        =     45°°  X  36°*  X  16  x  5  /      _  i  x     6 

24  X   16584000000  X  2i6\  6 

16  X  33        2  X  64  X  5\ 
-36"  ~2l6 1  =  °'13748  mch' 

5th,  These  4  equal  weights  remaining  on  the  beam,  what  is 
the  deflection  at  the  third  weight,  or  centre  ? 

In  equation  (281),  put  x  —  \l  =  cr,  r  =  3,  rx  =  4,  c  =  ^/; 


—       4500  X  36o3      j  /i8o  _        _  400X1 
6  x  16584000000(^72  432/8 

A  x  400       180       24^1       84 
i    V";   r^   rr^        ~rz"  *T~ 


24V 

12/4 


X  216         36         12/4        72 

6th,  The  same  4  weights  remaining,  what  is  the  deflection 
at  the  second  weight  ? 

Use  (281),  calling  r,  =  4,  r  =  2,  x  =  r^:  =  -|/,  ^  =  ^/; 

4500  x  3603     |/i8o  4oo\  i 

6  x  16584000000  (  ^^72"  '          ~  4/32/ 2~7 

400       i 80 


/th,  What  are  the  end  moments  due  these  4  weights  in  the 
same  position  as  above  ? 

Use  (283)  and  (284),  making  r  =  4,  c  =  \l  —  60,  W  = 
4,500; 


=  —750000  inch-pounds. 

45°°  X  3  °      ^Y3  x  20  —  2  x  9)  =  —600000  inch-pounds. 
12  36^6  / 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  21  7 

8th,  When  all  the  5  weights  are  on  the  beam  uniformly  dis- 
tributed as  above,  r  =  5,  c  —  \l  =  60,  W  =  4,500.  Then,  by 
(283)  and  (284), 

4500  x  360      30/44  9o\ 

M,  =        —  --  x  -g-^-g-  --  6  -  -^J       =  -78  7500  inch-pounds. 

4500  X  360       3°/3  \ 

M2  =  -       I2       -  X  -gU  X  30  —  2  x  II)  =  —78  75  oo  inch-pounds. 

9th,  The  4  equal  weights  of  4,500  pounds  still  occupying 
the  first  4  intervals  on  this  beam,  where  are  the  points  of 
contrary  flexure  ?  Here  we  have  M^  =  —  750,000,  M2  = 
—600,000,  W  =  4,500,  c  —  \l  =  60,  r,  =  4,  r2  =  d. 

These  values  put  in  (289)  give 

r  =  4-6595  or  1-1923. 

We  have  then,  rejecting  the  decimals,  r  =  i  or  4.    Hence  (288) 
becomes 

-750000  +  j.  x  45°°  x  *|fl  x  2 
=  -*  -  45oo(3  -  i  X  *  X  4  X  5)  =     74.806  for  r=  i, 


—  750000  4-  |  x  4500  x  s-jp  x  20 

= 


-  4500(0  -  i  X  i  X  4  X  5)  '  " 

ioth,  If  these  4  weights  occupy  the  last  4  intervals,  leaving 
the  first  vacant,  we  shall  have  M1  =  —600,000,  J/2  =  —750,000, 
r,  =  5,  r2  =  i,  c  =  £/  =  60,  PF  =  4,500  ;  so  that,  from  (289), 
we  find  r  =  4.80372  or  1.34442,  that  is,  4  or  i. 

These  values  placed  in  (288)  give 

x  =    84.706  for  r  =  i, 
x  =  285.194  for  r  =  4, 

which  accords  with  example  9th,  since  360  —  84.706  =  275.294, 
and  360  —  285.194  =  74.806. 


2l8  MECHANICS  OF   THE    GIRDER. 

nth,  When  all  5  weights  are  on  the  beam  at  equal  inter- 
vals, M!  =  M2  —  —787,500  by  example  8th.  Also,  c  —  \l  = 
60,  rt  —  5,  r2  —  o,  W  =  4,500.  From  (289),  we  find  that  r 
must  be  i  or  4  ;  and  therefore  (288)  gives,  as  the  points  of  con- 
trary flexure, 

x  =     y6f  for  r  =  i  , 

x  =  283^  for  r  =  4. 

The  sum  of  these  values  of  x  is  360,  as  it  should  be,  since  the 
load  is  symmetrical. 

1  2th,  Let  there  be  on  this  beam  weights  at  the  end  of  the 
second  and  third  intervals,  and  find  the  end  moments  and  points 
of  contrary  flexure.  We  now  have  W  —  4,500,  c  =  \l  =  60 
inches,  rI  =  3,  r2  =  i  ;  so  that  (290)  and  (291)  become 

M       —4500  x  360(6  i 

4f'=         -75—   -jgx  2x5-4x^(3x4x7-1x2x3) 

4.  -3-(i44  _  4)  j  =  -442500, 

210  ) 


-4500 


_  4)  J  =  _ 


Using  these  moments  in  (289),  we  find  r  =.  1.2460  or  4.2354; 
we  use  r  =  i  or  4.  Therefore,  from  (288),  x  =  79.254  or 
25I-°53>  which  are  the  points  of  contrary  flexure  sought. 

1  3th,  When  these  2  equal  weights  are  at  the  second  and  third 
points  of  division,  as  in  the  twelfth  example,  what  is  the  maxi- 
mum deflection  of  the  beam,  and  at  what  point  does  it  occur  ? 

Using  (281),  where  now  W  =  4,500,  c  =  60  =  J/,  rt  =  3, 
r2  =  i,  and  provisionally  putting  x  =  cr,  we  find  yr  ;  then, 
putting  r  +  i  for  r  in  the  value  of  yn  we  find  jv  +  i  J  and  then, 
making  Aj/  =  jj/r  +  r  —  yr  —  o,  we  obtain 


r3  —  6.722r2  •+•  9.1117*  -f  2.676  =  o, 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  2IQ 

from  which  we  easily  see,  as  was  suspected,  that  a  positive 
value  of  r  lies  between  2  and  3  for  a  maximum  y. 

Making,  therefore,  r  =:  2  in  equation  (281),  and  differentiat- 
ing with  respect  to  xy  then  putting  -^-  —  o,  we  find 

x  =  0.473914 
which,  substituted  in  (281),  r  being  2,  gives 

y  =  0.11553  inch,  a  maximum. 
At  centre,  y  —  0.11478  inch,  at  second  weight. 
At  \lt         y  =  0.09514  inch,  at  first  weight. 

87.  When  the  uniform  discontinuous  load  is  applied  at  equal 
consecutive  intervals,  the  first  weight  being  placed  at  no  inte- 
gral number  of  times  the  common  interval  from  the  left  end 
of  the  beam,  we  may  proceed  in  finding  the  deflection,  end 
moments,  and  points  of  contrary  flexure  as  in  article  20,  where 
r,  /•„  and  r2  need  not  be  integral,  but  where  the  differences, 
rl  —  r2,  r  —  r2)  rl  —  r,  each  denoting  a  number  of  weights, 
must  be  integral.  In  this  way  the  deflection  formulae  already 
established  in  this  chapter  for  full  intervals,  r,  rlt  r2,  being 
whole  numbers,  also  apply  to  the  case  now  under  consideration, 
where  r,  r,,  and  r2  have  the  same  fractional  part,  except  that, 
when  r2  is  negative,  its  value  is  less,  by  unity,  than  the  common 
decimal  part  of  r  and  /•„  as  before  shown. 

EXAMPLE  i.  —  Beam  fixed  horizontally  at  right-hand  end, 
simply  supported  at  left  end.  Length  =  360  inches  =  /,  c  = 
l/  =r  60  inches  ;  6  weights,  each  =  W  =  4,500  pounds,  applied 
at  the  intervals  \c>  c,  c,  c,  c,  c,  \c\  depth  and  cross-section  of 
I-beam  as  in  the  example  of  article  75,  where  the  moment  of 
inertia  of  section  —  I  =691,  and  E  =  24,000,000.  What  is 
the  deflection  at  the  centre  under  this  load  ? 


22O  MECHANICS  OF  THE   GIRDER. 

Using  equation  (292),  where  now  rz  —  —  J,  rz  —  5^-,  r  =  2^, 
and  x  ==  -|/  =  1 80,  we  find  central  deflection 

=        45°Q  X  360*       ( i  /3  .  I  .  6a 
12  x  16584000000(8^2  '  6  ' 


+  ^(44375  +  9  -  10.4583)  -  0.1771!  =  0.3993  i 

And  the  greatest  deflection  due  this  full  load  on  the  beam 
fixed  horizontally  at  the  right-hand  end  is  found  by  putting 
x  =  cr  provisionally  in  (292),  and  making  yr  +  l  —  yr  =  o  —  Ajj/. 
This  equation  indicates  a  value  of  r  between  f  and  f. 

Calling  r  =  J  in  (292),  and  putting  -2-  =  o,  we  find  x  = 

0.420777=  151.477  inches,  which  is  greater  than  c(r  +  i)  = 
6o(f  -j-  i)  =  150  inches,  an  inadmissible  result.  Hence  we  see 
that  the  approximate  equation  AJJ/  =  o  gave  r  too  small.  Now, 

calling  r  =  f  in   (292),  and  making  -^-   =  o,   we  find  .ar  = 

0.4174047  =  150.265  inches,  which  is  between  cr  and  ^(r  +  i), 
as  it  should  be. 

With  r  =  |,  and  x  =  0.4174047,  (292)  gives  greatest  deflec- 
tion jj/  =  0.41463  inch ;  while  at  the  centre  it  was  0.3993  inch. 
The  end  moment  in  this  case  removes  the  point  of  greatest 
deflection  180  —  150.265  =  29.735  inches  from  the  centre. 

The  end  moment  due  this  load  is  given  by  (293),  where 
rl  =  -y-,  rz  =  —•£,£  =  £/  =  60,  and  W  =  4,500  pounds ;  and 
it  is,  in  inch-pounds, 

4500  x  360(1         i 
M2  =         —^~ 

=  -1231875. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  221 

The  point  of  contrary  flexure  is  found  by  adding  equations 
(62)  and  (93),  and  equating  the  sum  of  the  second  members  to 
zero. 

Thus,  since  Ml  =•  o,  we  have 


r)/-  c(r,  -  r2)  (rt  +  r2  +  i)]  + 

-  ra)  (r  +  r2  +  i)  =  o,     (305) 


-c(r-ra)(r  +  r2  +  i)/  _ 

-  JJf-      (306) 
-  r)  -  <-(r,  -  ra)  (r,  +  r2  +  i)  + 


Making  x  =  re  provisionally  in  (306),  we  find  r  =  4.5353. 
Calling  r  =  |,  and  J/2  =  —1,231,875,  (306)  yields  the  point  of 
contrary  flexure 

x  =  0.7547174 

which  is  between  re  and  (r  +  i)c  (that  is,  between  O./5/  and 
J-J/),  though  very  close  to  the  former. 

If  both  ends  of  this  beam  are  free  under  this  load  of  6 
.equal  weights,  we  find  by  (232),  at  the  point  x  =  0.41  74O4/, 
y  =.  0.9676  inch. 

And,  by  (237),  the  deflection  due  M2  =  1,231,875  is  y  = 
—0.5530  inch,  which  added  to  0.9676  gives  y  —  0.4146  inch, 
as  found  by  (292)  above. 

88.  Continuous  Uniform  Load,  wfb,  on  Beam  fixed  Hori- 
zontally at  Both  Ends.  —  Take  the  examples  of  article  75, 
and  apply  to  the  deflections  there  found  the  effects  of  the  end 
moments  as  given  by  equation  (298). 

ist,  In  the  first  example  of  article  75,  for  beam  with  free 
ends,  the  deflection,  when  x  =  b  =  -J/  :=  120  inches,  and 
a  =  o,  was  found  to  be  y  =  0.23/1/1/1  inch. 


222  MECHANICS  OF   THE   GIRDER. 

Now,  by  (298),  the  effect  of  end  moments  on  the  deflection 
in  this  case  is 

_          75  X  360*        |  /j.   _  2  _   _£V_i_  _  i\ 
24  x  16584000000(^27       9       8i/\27       3/ 

/8        6        3\/i        A) 

Q-  )(-  —  -)>  =  —0.19392  inch. 

^27       9       8i/\9       3/j 

Therefore  the  deflection  sought  is 

y  =  0.23444  —  0.19392  =  0.04052  inch; 

the  left  third  of  the  1 5-inch  I-beam  bearing  75  pounds  to  the 
inch,  both  ends  being  fixed  horizontally. 

2d,  Again,  in  the  second  example  of  article  75  the  central 
deflection  =  0.24421  inch,  under  the  same  conditions.  If,  now, 
in  (298)  we  make  a  =  o,  b  —  \l  =  1 20  inches,  x  —  £/,  we  get 
the  effect  of  end  moments  on  deflection 

=          75  *  36°*         |  _  j*_  /i  __  i\ 
24   x  16584000000!       81  \8       2/ 

^  -f- jr~( )\~  ~ 0.20514  inch. 

Therefore  the  required  deflection  is 

y  =  0.24421  —  0.20514  =  0.03907  inch 
at  the  centre  of  the  beam  fixed  horizontally  at  both  ends. 

3d,  Applying  the  value  of  y  in  (298)  to  the  deflection  found 
in  the  third  example  of  article  75,  where  a  =  o,  b  =  ^/,  x  =  f /, 
we  find,  for  the  beam  with  fixed  ends, 

y  =  0.19176  —  0.17077  =  0.02099  inch. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  22$ 

4th,  The  greatest  deflection  due  75  pounds  per  inch  on  the 
left  third  of  this  I-beam  fixed  at  both  ends,  is  found  by  adding 
equations  (229)  and  (298),  and  in  the  resulting  equation  making 

-2-  =  o  when  a  =  o,  and  b  =  A/. 
dx 

This  gives  x  —  |/,  whence  the  greatest  deflection  y  = 
0.042199  inch  at  |/  from  the  left  end  of  the  beam,  which  is 
(|  —  |)/  =  Ty  beyond  the  end  of  the  load. 

5th,  The  end  moments  for  this  load  of  75  pounds  per  inch 
on  the  left  third  of  this  1  5-inch  I-beam  30  feet  long,  where 
/  rr  691,  E  =  24,000,000,  a  =  o,  b  =  ^l  ==  1  20  inches,  are 
given  by  equations  (296)  and  (297),  as  follows  : 


29    f  -  —  -  —  6  }  =  —330000  inch-pounds, 


M2  =  75  -  -  -    =  ~8ooo°  inch-pounds. 


6th,  With  these  values  of  Ml  and  M2,  equation  (300)  gives 
the  first  point  of  contrary  flexure, 

x  =  100.926  ±  37.229  =  63.697  inches, 

since  in  (300)  x  cannot  be  greater  than  (a  +  b)  =  (o  +  J/)  = 
1  20  inches. 

The  second  point  of  contrary  flexure  is  derived  from  (301), 
where  we  find  x  ==  260.69  inches. 

The  mode  of  procedure  when  only  one  end  of  the  beam  is 
fixed  horizontally  is  so  similar  to  that  just  exemplified  for  two 
fixed  ends,  that  further  examples  seem  to  be  unnecessary. 


224  MECHANICS  OF  THE   GIRDER. 


SECTION  4. 

Deflection  of  a  Girder  of  Variable  Cross-Section  in  Terms  of  the  Con- 
stant  Unit  Strain  upon  the  Extreme  Fibres  of  the  Section  ;  that  is, 
Deflection  of  a  Beam  of  Uniform  Strength.  End  Moments  for 
Fixed  Beams. 

89.  Economy  in  the  construction  of  built  beams  or  framed 
girders  requires  that  the  cross-sections  of  the  various  members, 
as  well  as  that  of  the  whole  structure,  should  be  proportioned 
to  the  greatest  strains  allowed  upon  the  sections  ;  and,  when  the 
dimensions  of  parts  are  so  adjusted,  it  is  clear  that  the  unit 
strain  of  tension,  compression,  or  bending  will  be  constant 
throughout  the  girder. 

The  complete  realization  of  this  condition  is,  for  obvious  con- 
siderations, probably  seldom  attained  ;  but  it  is  a  condition  so 
yearly  approximated  in  practice  as  to  require  examination  here. 

For  this  case  we  employ  equation  (186) ;  viz., 


\        17^, 


which  is  independent  of  /,  the  moment  of  inertia  of  the  cross- 
section,  4Jfid  in  which  B,  is  constant  for  a  given  load,  and  equal 
to  the  mean  of  the  unit  strains  upon  the  fibres  at  the  upper  and 
lower  surfaces  of  the  beam,  and  h  —  height  of  cross-section. 

90.  Deflection  of  Semi-Girder  of  Uniform  Height,  k, 
and  Uniform  Strength.  —  Using  the  notation  of  article  64,  as 
illustrated  by  Fig.  8,  and  integrating  (186),  with  the  sign  of 

E—  positive    for   the    semi-girder,  first,  with    the  condition 
dx* 

that        =  o  when  x  =  o, 


dx         h 


DEFLECTION,  END  MOMENTS,  ETC.,   FOUND.  22$ 


secondly,  7  =  0  when  x  =  o, 


h ' 

•••  y  =  -^-»  (307) 

which  is  the  deflection  at  any  point,  xy  of  the  semi-girder  of 
uniform  height  and  strength. 
If  x  —  /, 

D  =  ^m'  (3°8) 

which  is  the  deflection  at  the  free  end  of  the  semi-girder  of 
uniform  height  and  strength. 

It  may  be  observed  that  (307)  is  the  equation  of  a  parabola 
with  its  vertex  at  the  origin  of  co-ordinates. 

EXAMPLE.  —  Take  an  open-webbed  semi-girder  of  wrought- 
iron  whose  effective  height,  h,  is  20  feet  =  240  inches,  length,  /, 
=  50  feet  =  600  inches  ;  and  suppose  the  allowed  unit  strain 
in  the  top  chord  is  Ct  —  8,000  pounds  per  square  inch,  and  in 
the  bottom  chord  TI  =  10,000  pounds  per  square  inch.  Then 
calling,  as  we  may  do  without  sensible  error,  the  top  and  bot- 
tom chords  extreme  fibres  of  the  cross-section,  we  have 


Bi  =  \(Ci  +  T,}  =  9000.  (309) 

Take  E  =  25,000,000,  then 

9000  x  6oo2 
Deflection  at  free  end  =  D  =  240  x'a5Oooooo  =  °'54  mch' 


9000  x  3002 

Deflection  at  centre  =  —  -  -  -    -  =  0.135  mch. 
240  x  25000000 


226  MECHANICS   OF   THE   GIRDER. 

It  should  be  remembered  that  the  deflection  of  a  framed 
girder  due  to  its  first  full  load  is  likely  to  be  greater  than 
that  computed  by  these  formulas,  by  reason  of  the  yielding  of 
the  joints  and  probable  straightening  of  some  of  the  parts  in 
tension.  It  is  customary,  therefore,  in  computing  the  deflec- 
tion of  a  girder  under  its  first  loads  until  the  frame  becomes 
"set,"  to  take  E  ranging  from  15,000,000  to  20,000,000  for 
wrought-iron,  according  to  the  accuracy  of  the  joint  fittings  and 
general  workmanship  ;  afterwards  the  ordinary  value  of  E  may 
be  used. 

91.  Deflection  of  the  Semi-Girder  of  Uniform  Strength 
but  of  Variable  Height.  —  (a)  Let  the  semi-girder  be  like  either 
half  of  Fig.  64,  65,  67,  33,  34,  39,  or  83  ;  that  is,  let  it  slope 
uniformly  from  the  fixed  end,  whose  height  we  will  call  //„  to  the 
free  end,  whose  height  is  hv 

Then  the  height  at  any  point,  x,  is 

(310) 
.and 


Hence  (186)  becomes,  for  this  semi-girder, 

E(hQ  -  h,Y  #£=i 
2B*     "'  dh2       k' 


Integrating,  with  the  condition  that  -^  =  o  when  h 


. 

2HJ*  dh 

where  loge  denotes  the  Napierian  logarithm. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  22/ 

Again,  y  =  o  when  h  =  h1t 

h         \ 

*;  -  v  +  *" 


y  - 


which  is  the  deflection  of  the  uniformly  sloping  semi-girder  at 
any  point  where  the  height  is  h  ;  log  denoting  the  common 
logarithm,  and  the  girder  being  of  uniform  strength. 

Putting  for  //  in  (311)  its  value  as  taken  from  (310),  we  have 
y  in  terms  of  x  ;  thus, 


/&i/+(/&o— hi}x 

which  is  the  same  as  (311). 

If  the  semi-girder  of  uniform  slope  and  strength  comes  to  a 
point  at  the  free  end,  we  have  at  that  end  h0  =  o  =  h ;  and 
therefore  (311)  becomes 

7"»       7>% 

(313) 


which  is  twice  the  deflection  given  by  (308)  for  semi-beam  of 
the  same  length  but  of  the  uniform  height  hv 

When  //o  =  hi  =  //,  the  value  of  y  in  (311)  and  (312)  is  in- 
determinate, but  is  given  by  (307). 

EXAMPLE..  —  Length  of  semi-girder  /  =  50  feet ;  height  at 
fixed  end  =  20  feet,  at  free  end  10  feet ;  B^  =  9,000;  E  = 
25,000,000  pounds  per  square  inch.  What  is  the  deflection  at 
the  free  end  ?  //,  =  240  inches,  h0  =  k  =  120  inches,  /  =  600 
inches. 


228  MECHANICS  OF  THE   GIRDER. 


By  (3  1  1), 


=  0.6628  inch. 

(b)  Semi-Girder  with  Either  or  Both  Chords  Parabolic.  Open 
Frame.  —  First,  take  a  case  like  the  half  of  Fig.  63,  supposing 
the  top  chord  parabolic,  and,  as  in  all  these  cases,  the  members 
formed  as  for  a  semi-beam.  Let  /  =  length  of  semi-girder, 
//r  =  its  height  at  the  fixed  end,  hQ  =  height  at  free  end, 
and  h  =  variable  height.  Then,  by  equation  (136),  putting  for 
the  h  in  that  equation  h*  —  7z0,  and  adding  7z0  to  the  second 
member  for  our  present  case,  we  have,  /  also  being  put  for  J/, 


This  value  of  h  placed  in  (186)  gives,  after  reducing,  and 
making  m?  =  -  -  —  , 


-.^  =  _^ — .        (3:5) 

dx2       m2l2  —  x2 

Integrating   (315),  first   with  the    condition  -J-  =  o  when 
*  =  q, 

BJ  dx  *  ml  —  x 

where  loge  means  Napierian  logarithm. 

Integrating  again,  with  the  condition^  =  o  when  x  =  o, 


x)log(m/  +  x) 
(ml  —  x}\og(ml  —  x)  —  2ml\ogml~\, 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  22Q 


'-  *) 

(317) 


where  log  denotes  common  logarithm,  andjp  is  the  deflection  at 
any  point,  x,  of  the  semi-girder  of  uniform  strength,  and  of  the 
form  of  one-half  of  Fig.  63,  when  the  top  chord  is  parabolic. 

EXAMPLE.  —  Let   B^  =  9,000,  E  =  25,000,000,   /  =  600 
inches,  //,  =  240  inches,  k0  =  120  inches. 


If,  now,  x  =  /,  we  have  the  deflection  at  the  free  end  of  the 
girder,  from  (317), 

2.302585  x  9000  x  600 

y  =  ~  /-     (2°3-9)  =  0-59757 

25000000  x 


which  is,  as  it  manifestly  should  be,  less  than  the  deflection  just 
found  by  (311)  for  the  semi-beam  of  equal  length  and  depth 
of  ends,  but  of  uniform  slope,  and  greater  than  the  deflection  of 
semi-beam  of  same  length  and  uniform  depth  =  hu  found  by 
equation  (308). 

If  this  girder  comes  to  a  point  at  the  free  end  (that  is,  if  it 
is  the  half  of  the  parabolic  bowstring),  we  have,  in  (3  1  7),  k0  =.  o, 
m  —  i  ; 


+    (I  ~  *)lQg(/  -   X)    -    2/bg/],       (3l8) 

which  is  the  deflection  at  any  point,  x. 


230  MECHANICS  OF  THE   GIRDER. 

When,  in  (318),^  =  /,  we  have  the  deflection  at  the  free 
end  of  the  parabolic  semi-bowstring  ;  thus, 


~  I. 

D  = 


I.3862Q5 

which,  according  to  (313),  is  -  -  -  of  the  deflection  at  the 

free  end  of  the  semi-girder  of  same  length  and  height  at  fixed 
end,  but  sloping  uniformly  to  a  point. 

From  the  identity  in  the  form  of  equations  (136),  (137),  and 
(138),  and  from  the  manner  in  which  (317),  (318),  and  (319) 
have  been  derived  from  (136),  it  follows  that  the  deflection  of 
any  parabolic  semi-girder  of  uniform  strength,  whether  the  half- 
crescent,  or  the  half  double  bowstring,  may  be  found  from  (317), 
(318),  and  (319),  provided  we  make  /^  —  the  height  of  girder  at 
fixed  end,  and  hQ  =  its  height  at  the  free  end. 

(c)  Semi-Girder  with  Circular  Arc  for  Top  Chord.  Uniform 
Strength.  —  Let,  as  before,  //,  =  height  at  fixed  end,  hQ  =  height 
at  free  end,  for  a  girder  like  the  right  half  of  Fig.  63,  fixed  at 
the  vertical  plane  through  the  centre  ;  the  top  chord  being  now 
supposed  circular. 

If  ft  is  the  radius  of  the  circle,  the  height  of  the  semi-girder 
at  any  point,  xy  is  given  by  equation  (125), 

h  =  h,  -  hQ  +  hQ  - 

(320) 
h  —  hi  —  R  -f  /?cos0;  (321) 

0  being  the  arc  between  the  point  (x>  y)  of  equation  (125)  and 
the  fixed  end  of  the  girder. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND. 

Therefore  (186)  becomes 

E    dzy  i  ,       v 

—  7*  (322) 

2-Z?!  dxz       hi  —  R  -j-  R  cos  0 

But  x  =  R  sin  0, 


cos 0 dB  (       v 


Integrating  first  with  the  condition  that  -^  =  o  when  ^  =  o, 

dx 

we  have  (for   this   first   integration,  see  Price,  "Infinitesimal 
Calculus,"  vol.  ii.  p.  85),  after  reducing,  and  putting  a  =  cos  a, 


COS 


where,  as  usual,  loge  means  Napierian  logarithm. 

For  the  second  integration,  between  the  limits  o  and  7,  o 
and  0,  (324)  takes  the  form 


2J3.R  sin  a.        CQC  ^  +  6> 

2 

The  first  term  is  easily  integrated  thus : 

/     OcosOdQ  =  0sm0  -  \    sm0J0 

Jo  Jo 


.    (325) 


0sm0  4-  cos^  -  i 


232  MECHANICS  OF  THE   GIRDER. 

Integrate  the  second  term  also  by  parts,  according  to  the 
form 

fudv  —  uv  —  fvdu.  (326) 


Take  u  —  loge — -  =  loge  cos loge  cos  a  '     . 

CC  -f-  0  2  2 

2 

dfr  =  cos  0  //0,         /.    27  =  sin  0. 


sin 4  sm 

2 

tan 


cos cos  — - 

2  2 


Therefore  the  second  term  of  the  second  member  of  (325) 
becomes 

a  -  9 


cos 
a 


sin  0  loge ff  tan  ?—2  +  tan  ^±1\  sin 

o^«  +  0    2j  v      2  2  y 


4 

2 


But 


2  2  COS  C6  +  COS 

and  the  second  term  reduces  to 


since          CQS<*_±_0        J  cosa  +  cos0' 

2 

Now 


a  r      sing^      =  _a  p(cog8  +  cogg)  =  _atoge(cosa 
J  cos  a  4-  cos  0  J      cos  a  +  cos  0 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  233 

Whence,  finally,  the  integral  of  (325)  is 

a  -  0 
cos 

E 


sin  a  a  +  0 


!0 
, 
o 


...     y  _  0sin0  +  cos0  -  i 

-  0 


COS 


where  m^  =  0.4342945,  the  modulus  of  common  logarithms, 
log  ;  and  7  is  the  deflection  at  any  point,  x  =.  R  sin  0,  of  the 
semi-girder  having  its  top  chord  circular  and  bottom  chord 
straight,  like  the  truncated  bowstring. 

When  7/o  —  o  (that  is,  when  the  semi-girder  is  half  of  the 
common  bowstring  girder),  the  last  term  of  (327)  becomes  infi- 
nite for  a  =  —  cos  6,  which  is  the  case  if  x  =  I  —  length  of 
semi-girder,  and  sin  0  =  I  -r-  R. 

But  in  this  case  sin0  =  since;  and  (327)  is  easily  reduced 
to 

,  (328) 


which  is  the  deflection  at  the  free  end  of  the  circular  semi- 
bow-string  of  uniform  strength. 

EXAMPLE  i.  —  Semi-bowstring.     /  =  600  inches,  h^  =  240 
inches,  Bl  =  9,000,  E  =  25,000,000,  wrought-iron. 

...    R  =  /2  +  h*  =  870  inches  =72.5  feet, 


234  MECHANICS   OF   THE   GIRDER. 


sin(9  =  /-r-^  =  —  =  0.689655,      6  =  43°  36'  io".i5, 
87 

\ 

CoS  0  =  ^I_A  =        =  0.724138. 


In  arc,  0  =  43'6°2827r  =  0.761013, 


1  80 


a  =  -^-^  —  =  --  &  =  —0.724138  =  cos  a  =  —  cos0, 
fl  87 

/.     a  =  180°  -  43°  36'  io".is  =  136°  2/49^5, 


Therefore  the  deflection  at  the  free  end  of  this  semi-bow- 
string of  uniform  strength  is,  by  (328), 

2  x  9000  x  870 
y  =  -  25000000  —  (°'524836  +  0.724138  -  i  +  1.145371) 

=  0.71745  inch, 

which  is  a  little  less  than         2         X   1.08  =  0.7486  inch  = 

deflection  at  free  end  of  parabolic  semi-bowstring,  by  (319). 
And  this  should  be  so,  since  the  top  chord  of  the  parabolic 
girder  lies  just  below  that  of  the  circular  bowstring  of  the 
same  central  height  and  same  span. 

EXAMPLE  2.  —  Semi-girder,  truncated  bowstring,  circular. 
/  =  600  inches,  //,  =  240,  hQ  =  120,  B*  —  9,000,  wrought-iron  ; 
E  =  25,000,000  ; 

.-.     R  =  /2  +  ^'  ~  h°>*  =  130  feet  =1560  inches, 
2  Ax  -  Ao 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  235 

Use  equation  (327). 

sin  6  =  I  -i-  R  —  -fa, 

0  =  22°  3/  1 1"-5  =  22>6lg986^  =  0.39479  m  arc. 

IoO 

2O  —    130  II 

0050  =  0.923077,         0  =  cos«=        I3Q  -  —  5 

sina  =  0.532939,         a  =  180°  -  32°  i2'is".3  =  147°  47' 44^7 ; 
£L±^=  8S°i2'28".i,         £LH_?=  62035'i6".6; 

2  2 

2  X  QOOO  X   1560 

•'•  •>  =  -  25000000    (0.151842+0.923077  - 1+0.455710) 

=  0.596003  inch, 

which  is  the  deflection  at  the  free  end,  and  is,  as  was  to  be 
expected,  a  little  less  than  that  found  by  (317)  for  the  parabolic 
semi-girder  of  the  same  length  and  end  heights. 

92.  Equations  (327)  and  (328)  apply  also  to  the  double  cir- 
cular bowstring,  truncated  or  otherwise,  provided  the  radii  of 
the  two  curves  are  the  same.     But  when  these  radii  are  differ- 
ent,  we    may,   without    sensible    error,   employ   the   equations 
(317),  (318),  and  (319),  deduced  for  the  deflection  of  the  para- 
bolic semi-girder  of  uniform  strength,  and  applicable  to  all  the 
cases,  including  the  crescent  and  the  double  bow;  the  computed 
deflection  being  always  a  little  greater  than  that  due  the  cir- 
cular semi-girder  of  the  same  end  heights  and  span. 

93.  Deflection  of  the  Girder  of   Uniform  Strength  sup- 
ported at  Both  Ends,  either  Fixed  or  Free,  and  the  Height 
of  the   Girder  being  either   Uniform   or  Variable.  —  Since 
the  deflection  of  a  girder  may  be  defined  as  the  difference  of 
level  between  the  position  of  any  one  of  its  points  before  bend- 
ing and  the  position  'of  the  same  point  after  bending,  under 


236  MECHANICS  OF   THE    GIRDER. 

the  given  load,  it  follows  that  the  formulae  already  established 
for  the  deflection  of  the  semi-girder  of  uniform  strength  also 
apply  to  the  present  case,  provided  we  take  the  origin  of  co-or- 
dinates in  the  neutral  axis  at  the  centre  of  the  span,  and  call 
y  positive  upward,  and  write  \l  for  /;  /  being  the  length  of 
the  girder  in  all  cases,  and  the  neutral  axis  taken  horizontal. 

EXAMPLES.  —  Take  an  open  webbed  girder  of  wrought-iron, 
height  at  the  centre  —  25  feet  =  300  inches,  span  =  200  feet 
=  2,400  inches  ;  therefore  kt  =  300,  \l  —  1,200.  Let  B^  — 
%(d  +  Tt)  —  9,000,  E  =  25,000,000.  What  is  the  deflection 
at  the  centre  ? 

EXAMPLE  i.  —  Height  uniform  =  k  =  /i,  —  300;  therefore 
central  deflection  is,  by  (308), 

OOOO    X    I2OO2 

D  =  --  =  1.728  inches. 
25000000  X   300 

EXAMPLE  2.  —  Truncated  circular  bowstring,  h^  =  300, 
k0  =  1  80  =  end  height,  /2r  —  h0  =  120. 

R  =   (VY  +    (*.  "   h0Y   =    I2002  +    120- 


Use  equation  (327). 


= 
2    X    120 


sin  0  =  ^  =  —  =  0.198020,        6  =  1  1°  25'  i6".3  ; 
R       5°5 


^0.980:98. 


In  arc,  0  =  TT  =  0.199338. 

1  80 

7  7? 

cos  a  =  a  —  -^     -  =  —0.950495. 
K 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  237 

«  +  0)  =  86°39'3i".i5,   a  =  180°  -  18°  6'  14"  =  161°  53'  46". 
a  -  0)  =  75°  14'  14".  85. 


=  0.039473,        \og  =  -0.2218488. 

cosM^l  sinj?       cosK«-0)  =  -0.408267. 

&cosj(a  +  6»)  sin  a       cosi(a  +  0) 

—  (0.408267  —  0.2218488)  =  0.407994. 


25000000 
=  1.86169  inches. 


( 


EXAMPLE  3.  —  Truncated  parabolic  bowstring,  equation  (317). 
—  300,  7/o  =  1  80,  x  —  -|-/  =  1,200,  m*  —  -  —  IJL—  =  2.5. 

fil  "O 


=  1897.366,    m(\l)  +  x  —  3097.366,    w(-|/)  —  x  =  697.366. 
log|w/=  3.2781512,         log  (-|^/+  JP)  =  3.4909925, 
log  (^ml  —  x)  —  2.8434608. 

Therefore  (317)  becomes 

2.302585  x  9000  x  1200  , 

y  =  —  5  —  2_2  -  Z  -  (10812.88  -|-  1982.93  —  12439.70) 
7       1.58114  X  120  X  25000000  v 

=  1.86695  inches. 

EXAMPLE  4.  —  Chords  of  uniform  slope,     h^  =  300,  hQ  = 
1  80  —  //,  \l  =  1,200.     Use  equation  (311). 


y  =  ~  ' '  7~~~  "  """°  <  3°°  —  i8o[  2.302585  log 
-  "  --->2  (       \ 


_  2    X    9OOO    X    I2OO2 

25000000  x  i202  \^  ~  \    ^     u'u~  °  180 

=  2.0197  inches. 


238  MECHANICS  OF   THE    GIRDER. 

EXAMPLE   5.  —  Circular  bowstring.     \l  =  1,200,  /^  =  300, 
hQ  =  o.     Use  equation  (328). 


-  =  212.5  feet  =  255°  inches. 


sin0  =       =  0470588,         0  =  28°4'2i". 
R 

7?         Jt 

cos0  =  —  —  —  -  =  0.882353.         In  arc,  0  =  0.501346. 
I? 

cos  a  =  —  cos#  =  a  =  —0.882353. 

a=  180°-  28°4'2i"=  151°  55'  39". 

*(•  -  <0  =  6I-SS'  39",     ^2[cos:(t:,)]2  =  -0-57573x8- 
=  0.23593. 


2    X    QOOO    X    -  ,  ,- 

y  = —  (0.2^91  +  0.882353  — 

J  25000000 

+  2.302585  x  0.882353  X  0.5757318)  =  2.36477  inches. 

EXAMPLE  6.  —  Parabolic  bowstring.    \l  —  1,200,  h*  =  300, 
By  equation  (319), 

1.^8620^     X    QOOO    X     I2OO2 

D  =  — =  2.39552  inches. 

25000000   X   300 

EXAMPLE   7.  —  Girder   sloping   uniformly   from   centre  to 
ends.     \l  =  1,200,  hr  —  300,  k0  =  o. 
By  equation  (313), 

2    X    9000    X     I2002 

D  = =  3.4^6  inches. 

25000000  X   300 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  239 

EXAMPLE  8.  — Parabolic  crescent.  //,  =  300,  h0  =  o,  \l  = 
1,200. 

The  deflection  in  this  case  must  be  the  same  as  that  in  the 
sixth  example,  for  the  parabolic  bowstring. 

/.     D  =  2.39552  inches. 

EXAMPLE  9.  —  Girder  like  Fig.  5  3,  sloping  uniformly  from 
centre  to  ends.     //x  —  300,  7z0  =  o,  \l  —  1,200. 
Deflection  the  same  as  in  example  7,  viz., 

D  —  3.456  inches. 

EXAMPLE  10.  —  Girder  like  Fig.  66,  polygonal. 
Find  the  deflection  for  each  part  having  a  uniform  slope, 
separately,  and  add  the  results  for  the  total  central  deflection, 
after  correcting. 

Take  /it  =  300  at  Z4,  and  h0  —  240  at  Z6,  the  quarter-section. 
Then  \l  =  600,  and  equation^  11)  gives  the  deflection  at  Z6, 
thus, 

2  x  9000  x  600% 
y  =    25000000  x  60*  l>o  -  240(2.302585  log f£J  +  O] 

=  0.46408  inch. 
Similarly,  for  the  end  quarter,  equation  (313)  gives 

2  x  9000  x  6oo2 
=    25000000  x  240    '  =  i. 08  inches. 

But,  before  adding  these  results,  we  must  find,  as  in  article 
67,  how  much  the  free  end  of  the  semi-beam  is  deflected  by 
reason  of  the  bending  of  the  part  between  Z"4  and  ^6 ;  that  is, 
we  must  add  to  0.46408  the  quantity  \l  X  tan  a  =  600  tan  a 

& 
dx 


240  MECHANICS  OF  THE   GIRDER. 


From  (311), 

=  tana  =   _.1  ,  .log.|.  =  0.0016066, 
*i)        "4 


600  X  0.0016066  =  0.96398  inch, 
/.     Total  deflection  =  0.46408  -f  0.96398  -f-  1.08  =  2.50806  inches, 

which  is  greater  than  the  deflection  found  in  example  6,  for 
the  parabolic  bowstring  ;  and  it  will  be  found  that  although  the 
girder,  Fig.  66,  is  deeper  at  the  quarter-points  than  the  para- 
bolic bow  of  example  6,  yet  at  the  }  and  f  points  the  latter  is 
the  deeper. 

In  like  manner  may  we  proceed  in  all  cases  of  irregular 
forms,  whether  there  be  two  or  more  changes  of  slope  ;  but,  in 
general,  we  may  use  the  formulae  already  found  for  regular 
forms,  with  sufficient  accuracy,  always  choosing  the  one  most 
fitting  for  the  case  in  hand.  f 

94.  We  may  arrange  the  results  found  in  these  examples 
according  to  the  amount  of  the  deflection,  and  thus  the  more 
clearly  perceive  the  effect  of  form  upon  the  bending  of  girders 
of  uniform  strength.  All  the  girders  here  represented  are  200 
feet  in  length  if  supported  at  both  ends,  or  100  feet  long  if 
semi-girders  ;  the  deflection  being  the  same  in  either  case. 

Since  in  all  the  formulas  the  deflection  varies  directly  as 


2*  —  K^i  +  T*\  we  may  find  the  deflection  of  girders  of  the 
E  E 

same  dimensions,  but  of  other  material  than  wrought-iron,  by 
substituting  for  E  the  proper  value  taken  from  Table  II.,  and 
for  Ct  and  2"x  the  allowed  unit  strain. 

If,  for  pine,  Tt  =  1,200,  Ct  —  552,  E  —  1,460,000,  then  B^ 

7?  /? 

=  876,  and  —  -  =  0.0006,  but  for  wrought-iron  -r  =  0.00036  ; 

hence,  for  a  girder  of  uniform  strength  and  of  given  span  and 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  24! 

height,  the  deflection,  if  the  material  is  pine,  will  be  five-thirds 
of  the  deflection  were  the  material  wrought-iron  ;  that  is,  allow- 
ing Ct  and  Ji  the  above  values. 

If  the  compressed  chord  be  of  pine,  Cl  =  552,  and  the 
other  of  wrought-iron,  Tt  =  10,000,  and  if  E  =  13,230,000  = 

ry 

4(25,000,000  +  1,460,000),  we  have  B^  =  5,276,  — -  =  0.0004.. 

R 

Hence  a  combination  of  pine  and  wrought-iron  gives  a  deflec- 
tion -4-  =  —  times  that  due  wrought-iron  alone,   with  these 
3-6        9 

unit  strains. 

Were  the  compressed  chord  of  cast-iron,  for  which  Ct  — 
15,000,  while  the  other  chord  is  of  wrought-iron,  Tl  =  10,000, 
and  E  =  \(2 5, 000,000  +  12,000,000)  =  18,500,000,  we  should 

n 

have  BI  —  12,500,  —  —  0.00067567,  and  the  deflection  would 
E 

be  1.877  times  that  of  the  girder  of  same  size  in  wrought- 
iron. 

95.  By  inspecting  the  following  table,  we  see  that  for  open 
girders  of  the  same  central  height,  same  length,  and  of  uniform 
strength,  the  total  deflection  is  NEARLY  in  the  inverse  ratio  of  the 
areas  of  the  figures  of  the  girders. 

This  is  exactly  the  ratio  of  the  deflections  in  case  of  the 
girder  of  uniform  height  and  of  that  sloping  uniformly  to  a 
point :  viz.,  ratio  of  areas,  \ ;  ratio  of  deflections,  \.  We  may, 
therefore,  without  appreciable  error,  employ  this  principle  in 
finding  the  total  deflection  of  open  girders  of  uniform  strength 
and  variable  height. 


242 


MECHANICS   OF   THE   GIRDER. 


EXAMPLES.  —  DEFLECTION  OF  OPEN  WEBBED  GIRDERS  OF  UNIFORM  STRENGTH. 

Length  =  /  =  200  feet,  central  height  =  /Zj  =  25  feet. 

MATERIAL. 

Wrought- 
Iron. 

Pine. 

Wrt.-Iron 
and  Pine. 

Wrt.  and 
Cast  Iron. 

Equa- 
tion. 

E 

9,000 
25,000,000 

876 
1,460,000 

5,276 
13,230,000 

12,500 
18,500,000 

Form. 

Description. 

Def.,  ins. 

Def.,  ins. 

Def.,  ins. 

Def.,  ins. 

100                   \ 
25                            25 

25                                   I 

1. 

S 

Uniform      ) 
height  J 

1.728 

2.88 

1.92 

3-24 

(308) 

2. 

25    ^  ^"""^         i 

One  chord  | 
circular  f 

1.86169 

3.1028 

2.0685 

3494 

(327) 

^===^ 

3. 

~^===^      "751 

One  chord  ) 
parabolic  ) 

1.86695 

3.III6 

2.0744 

3.504 

(317) 

^s======^ 

4. 

25                              15] 

Uniform      ) 
slope  ) 

2.0197 

3.366 

2.2441 

3-791 

(311) 

25             IB] 

5. 

Tf^          <^" 

Circular       \ 
Bowstrfhg  ) 

2.36477 

3.9413 

2.6275 

4438 

(328) 

25        ^  — 

€. 

25                    -^^^ 

Parabolic    ) 
curves  ) 

2.39552 

3.9925 

2.6617 

4.496 

(319) 

25       ^  — 

7. 

25         ~^-~^^^ 

Uniform      ) 
slope  } 

3456 

5.76 

3-84 

6.48 

(313) 

2T^—  ^_^ 

25  ^  -**^" 

25          ~^^Z=~ 

8. 

^S^^ 

Polygonal    ) 
chord  ) 

2.50806 

4.1801 

2.7867 

4703 

(3") 

DEFLECTION,   END   MOMENTS,   ETC.,   FOUND. 


243 


We  give  below,  the  deflections  of  girders  of  wrought-iron 
for  the  eight  cases  just  tabulated,  but  now  computed  by  this 
Method  of  Areas :  — 


No. 

Area  of 
One-Half  Girder. 

Deflection. 

Deflection  by 
Formulae. 

I 

2500  square  feet. 

1.7280  inches. 

1.72800  inches. 

2 

2167       " 

1.9931 

1.86169    " 

3 

2167       " 

1.9938        « 

1.86695    " 

4 

2000         « 

2.1600       " 

2.01970     " 

5 

1737         " 

2.4871        « 

2.36477     " 

6 

1667         " 

2.5920       " 

2.39552    " 

7 

1250         "             " 

3.4560       « 

3.45600     " 

8 

1625         " 

2.6585 

2.50806     " 

The  deflections  of  such  girders  as  those  shown  in  Figs.  19, 
20,  29,  30,  31,  32,  33,  34,  39,  40,  53,  54,  55,  etc.,  are  therefore 
easily  found  by  the  method  of  areas. 

It  should  be  noticed  that  in  the  preceding  table  of  deflec- 
tions of  the  same  girder  in  different  materials,  a  factor  of  safety 
equal  to  10  has  been  allowed  for  pine,  while  5  is  the  factor 
allowed  for  wrought  and  for  cast  iron. 

The  modulus  of  elasticity  for  cast-iron,  E  =  12,000,000,  is 
so  small,  that,  in  spite  of  its  large  resistance  to  compression, 
d  =  15,000,  the  open  beam  made  of  wrought  and  cast  iron, 
and  of  uniform  strength,  has  greater  deflection  than  that  of 
wrought-iron  alone,  and,  indeed,  greater  than  that  of  pine 
alone,  with  the  low  unit  strain  here  allowed. 

96.  Finally,  if  the  beam  be  of  uniform  strength,  but  have  a 
continuous  web,  the  formulae  already  deduced  for  girders  of 
uniform  strength  and  of  open  web  may  be  employed  by  assign- 
ing to  BI  its  proper  value  derived  from  Table  II. 


244  MECHANICS  OF   THE   GIRDER. 

EXAMPLE  i.  —  Plate  girder  of  uniform  strength  and  uniform 
height,  wrought-iron.  Take  the  length  /  =  50  feet  =  600 
inches,  height  h  =  5  feet  =  60  inches  ;  the  girder  being  sup- 
ported at  both  ends. 

By  Table  II.,  B  =  42,000  —  breaking  unit  strain  for  plate 
beams.  Allowing  a  safety  factor  of  5,  we  have  Bl  =  8,400  ; 
and  calling  E  =  25,000,000,  and  putting  £/  for  /  in  equation 
(308),  there  results  the  central  deflection, 


D  =  <  30°'     =  0.504  inch. 

25000000  X   60 

EXAMPLE  2.  —  Take  a  plate  girder  of  the  same  length,  50 
feet,  and  same  central  height,  5  feet,  but  sloping  uniformly 
from  centre  to  ends,  where  the  height  is  2  feet. 

Then,  if  the  girder  is  of  uniform  strength,  we  have,  from 
equation  (311), 


,  x  8400  x  [6o  _       (  , 

25000000  X  (60  —  24)*  L 

=  0.65376  inch. 

EXAMPLE  3.  —  Cast-iron  beam  of  uniform  strength,  and 
height  h  =  3  feet  =  36  inches,  -|-/  =  6  feet  =  72  inches. 
Take  Bt  —  *&%&  =  7,650,  E  =  17,000,000  (Table  IL). 

Then,  by  (308), 

7650  x  722 
D  =  —  —  -  ---  -  -  7  =  0.0648  inch. 

17000000  X  36 

EXAMPLE  4.  —  If  this  cast-iron  beam  of  uniform  strength 
slope  uniformly  from  centre  to  ends,  where  h  =  12  inches,  then, 
by  (311), 

-7^6  -  12(2.302585  logff  +  i)] 


=  0.0876  inch. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  24$ 

EXAMPLE  5.  —  Oak  beam  of  uniform  strength  and  height. 
Take  \l  —  1  20  inches,  h  =  1  8  inches,  B^  =  -L^°—  ,  E  —  2,  1  50,000  ; 
then  (308)  gives 

1060    X     120* 


D  = 


2150000  x  18 


EXAMPLE  6.  —  If  this  oak  beam  of  uniform  strength  slope 
uniformly  from  centre  to  ends,  where  h  =  12  inches,  then,  by 

(311), 

2  X  1060  X  I202 

^18  -  ^(2.302585  log  it  +  1)] 


2150000  x  (12  -  ts) 

=  0.4474  inch. 

EXAMPLE  7.  —  Beam  of  Bessemer  hammered  steel,  uniform 
strength.  \l  •=.  72  inches  ;  height  at  centre,  /it  =  20  inches, 
at  ends,  //0  =  10  inches.  Take  B,  =  l^d&l  =  25,616,  E  = 
31,000,000  (Table  II.). 

Then  deflection  at  centre  is,  from  (311), 

2    X    25616    X    722 

'  =  31000000(10  -  2o)'t30  -  IO(2-3°2S85  logM  +  i)] 
=  1.1196  inches. 

For  same  beam  of  wrought-iron, 

2  x  9000  x  722 
y  =  25000000(10  -  2o)»l>0  ~  ^(2-302585  log  fg  +  i)] 

=  0.4878  inch, 

which  is  less  than  half   the  deflection  of   the  same  beam  in 
steel. 

But  if  we  suppose  this  beam  to  be  of  rectangular  cross- 
section,  and  to  bear  a  concentrated  weight,  W,  at  its  centre, 
where  the  height  is  //,  =  20  inches,  and  the  thickness  b  =  2 


246  MECHANICS  OF  THE   GIRDER. 

inches,  the  length  being  /  =  144  inches,  then,  from  equations 
(46)  and  (160),  we  have  moment  at  centre, 

M  =  \Wl  =  %Bbh2  =  %£M2  for  safety, 


3  M4  ' 

W =  3.7037  x  25616  =  94874  pounds  for  steel, 

W  —  3.7037  X  9000  =  33333  pounds  for  wrought-iron. 

Hence,  under  the  assumed  unit  strains,  the  steel  beam  bears 

=  2.8462  times  the  weight  at  the  centre  of  the  wrought- 
9000 

iron  beam  of  the  same  dimensions,  while  the  deflection  of  the 

steel  beam  is  ^-^    -  =  2.2953  times  that  of  the  wrought-iron 
0.4878 

beam ;  that  is,  what  is  shown  in  all  the  formulae,  the  weight 
W  varies  directly  with  the  unit  strain  Blt  while  for  the  same 
unit  strain  the  deflection  varies  inversely  as  the  modulus  of 
elasticity,  E. 

Therefore  in  the  present  case,  so  far  as  deflection  is  con- 
cerned, the  advantage  of  steel  over  wrought-iron,  under  same 

load,  is  -  p —  =  — ,  which  is  the  simple  ratio  of  the  moduli 

of  elasticity. 

97.  The  thickness,  b,  of  a  continuous  webbed  girder  of 
uniform  strength  at  any  rectangular  section  of  given  height, 
//,  may  be  found,  in  general,  by  equating  the  moment,  M,  due 
the  external  forces,  to  the  moment  of  resistance,  R,  of  the 
internal  forces  of  the  beam  at  the  given  section,  and  solving 
with  respect  to  b. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  247 

For  a  beam  of  rectangular  cross-section,  bearing  a  con- 
centrated load  at  its  centre,  equations  (45)  and  (160)  give 
M  =  Wx  =  IBM2-  BI  —  allowed  unit  strain. 


(329) 


where,  if  the  height,  //,  be  uniform,  b  varies  as  x\  making 
the  horizontal  projection  or  ground  plan  of  each  half  of  the 
beam  a  triangle  with  a  vertex  at  the  end  of  the  beam,  where 

b  =  x  =  o,  and  a  base  at  the  beam's  centre,  where  b  =  -  -  . 


EXAMPLE.  —  Oak   beam  of  uniform   strength,  and   height 
h  =  15    inches,    length  =  15   feet,  weight   applied   at   centre 

—  W  =  4,000  pounds  allowed  unit  strain  =  -  =  1,060 

10 

pounds  per  square  inch.     What  must  be  the  thickness  of  this 
beam  at  the  centre  ? 

Here  x  =  \l  —  90  inches, 

/.    b  =  3  x  4°°°  X  90  =  inches< 

1060  x  i2 


It  must  be  remembered  that  wherever  the  moment  becomes 
zero,  causing  b,  the  thickness  of  the  beam,  to  vanish  by  the 
formulae,  we  must,  nevertheless,  have  at  all  such  points  suffi- 
cient material  to  resist,  with  the  proper  margin  of  safety,  the 
shearing-strain  which  may  there  be  developed,  and  the  re-action 
of  the  supports. 

In  this  example  the  shearing-strain  at  each  end  of  this  beam 

4000 
is  —  —  =  2,000  pounds.    Now,  by  Table  L,  the  ultimate  resist- 

ance to  shearing  is,  for  oak,  across  the  grain,  4,000  pounds  ; 


248  MECHANICS  OF  THE   GIRDER. 

one-tenth  of  which  is  400  pounds,  to  be  allowed  to  each  square 
inch  of  the  vertical  section  at  each  end. 


Therefore  -  =  5  square  inches  of  section  at  least  the 
400 

beam  must  have  at  each  end;  that  is,  the  depth  being  15 
inches,  the  thickness  is  \  inch.  But  there  is  another  consid- 
eration to  be  attended  to  ;  viz.,  the  bearing-surface  at  the  ends 
must  be  sufficient  to  resist  with  safety  and  permanence  the 
pressure  coming  upon  it. 

This  beam  as  now  estimated  is  \  inch  thick  at  each  end, 
and  4.53  inches  at  its  centre.  Hence  it  must  have  8.903  inches 
of  its  length  at  each  end  upon  the  support,  in  order  to  secure  a 
bearing  of  3^  square  inches,  required  for  2,000  pounds  with  an 
allowed  unit  strain  of  600  pounds  to  the  square  inch,  in  com- 
pression. 

Again,  a  beam  so  thin  at  the  ends  would  lack  lateral  stiff- 
ness unless  it  were  walled  in. 

In  practice,  therefore,  even  when  it  is  desired  to  use  the 
least  material  possible,  it  is  customary  to  make  those  parts  of  a 
beam  which  theoretically,  or  rather,  by  formula,  are  almost 
nothing,  of  such  size  as  a  just  regard  to  all  these  require- 
ments, as  well  as  to  the  good  appearance  of  the  structure,  may 
demand. 

Let  it  not  be  inferred  that  theory  and  practice  are  at  vari- 
ance here,  for  such  is  not  the  case.  The  equations  which 
determine  the  thickness  of  the  beam  do  not  pretend  to  take  into 
the  account  all  the  conditions  affecting  the  sufficiency  of  the 
beam  for  its  purpose.  And  hence  the  theory  is  not  complete 
till  the  modifying  conditions  are  introduced. 

98.  If  the  beam  of  uniform  strength  be  loaded  uniformly 
with  w  units  of  weight  to  the  unit  of  length,  we  have,  from 
equations  (49)  and  (160), 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  249 


putting  BI  for  B,  and  the  cross-section  being  rectangular  ; 


which  is  the  thickness  of  the  beam  at  any  point,  x,  measured 
from  the  end.  When  h  is  constant,  (330)  is  the  equation  of  a 
parabola ;  the  vertex  being  at  the  end  of  the  beam. 

Thickness  at  end      =  b  =  o.  x  =  o. 

Thickness  at  centre  =  b  =   ^W    .         x  —  \L 


Horizontal  projection,  two  parabolas. 

EXAMPLE.  —  Oak  beam,  uniform  strength.     Height  uniform 


=  h  —  15  inches,  length  /  =  180  inches,  B,  =      —  I)O6o, 

10 

8OOO  A  '  1 

w  =  -     -  =  44!  pounds  per  inch. 
1  80 

Then  thickness  at  centre  is 

x  400  X  i8o2 


•*  x  400  x  iocr  .    , 

— 2 2 =  4.53  inches. 

4  X  9  X  1060  x  is2 


99.  If  the  cross-section  of  the  beam  of  uniform  strength  be 
of  either  form,  Fig.  91,  then,  by  assigning  values  to  three  of 
the  dimensions,  //,  //„  b,  biy  we  may,  from  equation  (161)  and  the 
equation  expressing  the  moment  due  the  given  load,  find  the 
fourth  dimension  of  the  cross-section,  which,  therefore,  becomes 
known  at  every  point. 

In  like  manner  may  we  determine  any  one  dimension  of  any 
cross-section  whose  moment  of  resistance,  R,  is  known. 


250  MECHANICS  OF  THE   GIRDER. 

EXAMPLE.  —  Take  a  tubular  plate  girder  of  the  dimensions 
given  in  example  i,  article  96 ;  viz.,  /  =  50  feet,  h  =  5  feet,  Bt 
=  8,400,  uniform  strength  and  height.  Cross-section  as  in 
Fig.  91,  where  let  b  =  12  inches,  £x  =  \\\  inches;  the  side 
plates  being  J  inch  thick  each,  h  =  60  inches. 

From  (49)  and  (161),  we  have 


equal  to  58  inches  if  w!  =  123,508  pounds,  the  total  uniform 
load  on  beam,  and  x  =  ^/  =  300  inches. 

At  the  centre,  therefore,  the  top  and  bottom  plates  must 
have  the  thickness  of  i  inch  each ;  while  at  the  ends,  where 
x  —  o,  (331)  gives 

hi  =  hi )3  =  1.02174^  =  61.3044  inches, 

\«-W 

which  renders  h  —  h^  —  —1.3044  inches  negative,  showing 
that  the  cross-section  of  the  side  plates  is  more  than  sufficient 
at  the  ends  to  resist  the  moment. 

We  may  find  at  what  distance  from  either  end  of  this  beam 
the  top  and  bottom  plates  begin  to  be  needed,  by  putting  /^  = 
h  =  12  in  (331),  and  finding  x.  This  gives  x  =  69.19  inches, 
for  which  the  side  plates  alone  are  sufficient  if  properly  braced 
laterally.  Now,  the  shearing-strain  at  each  end  of  the  beam 
supporting  this  load  is  J  X  123,508  —  61,754  pounds;  and, 
calling  the  allowed  shearing-strain  8,000  pounds  to  the  square 

inch,  we  require  -o =  7.72  inches  in  cross-section  of  the  two 

plates,  whereas  we  have  2  X  f  X  60  =  45  square  inches.    But 


DEFLECTION,  END  MOMENTS,   ETC.,  FOUND. 

in  order  to  have  sufficient  bearing-surface  on  the  abutments, 
allowing  the  iron  to  bear  8,000  pounds  to  the  square  inch  in 
compression  also,  the  beam  must  be  supported  for  at  least 

7  72 
'  3  =  10.3  inches  of  its  length  at  each  end. 

2   X    g^ 

The  semi-girder  of  uniform  strength  and  continuous  web  is 
to  be  treated  in  the  same  manner  as  the  girder  just  considered 
when  we  seek  its  variable  cross-section. 

100.  Beam  of  Uniform  Strength  fixed  Horizontally  at 
Both  Ends.  —  By  definition  the  beam  of  uniform  strength  is 
equally  efficient  at  all  sections  to  resist  the  strains  generated 
by  the  external  forces.  Hence,  when  this  beam  is  horizontally 
fixed  at  both  ends,  and  loaded  with  a  concentrated  or  with  a 
continuous  load,  the  points  of  contrary  flexure  are,  for  any  style 
of  beam  or  girder,  practically  midway  between  the  centre  of 
gravity  of  the  load  and  the  ends  of  the  girder ;  since  there  is 
as  much  reason  for  their  being  on  one  side  of  this  midway 
point  as  there  is  for  their  being  upon  the  other  side  of  it,  and 
no  more.  And  the  beam  of  uniform  strength  is  such  only  with 
reference  to  a  particular  mode  of  loading.  That  is,  if  the  unit 
strain  is  uniform  throughout  the  girder  for  a  given  position  of 
the  load,  a  change  in  the  position  of  the  load  causes  a  change 
in  the  relative  values  of  the  total  strains  in  the  members  or  parts 
of  the  girder,  and  therefore  a  change  in  the  unit  strain  on  each 
member,  if,  as  is  assumed,  the  cross-sections  of  the  members  be 
not  changed. 

In  general,  we  have  for  any  girder,  from  equations  (184) 
and  (187), 

Moment  due  internal  forces,  Mx  =  ^^  =  2jgltS>2,     (332) 

h  h 

where  Bv  =  allowed  unit  strain  in  bending,  5  =  area  of  any 
cross-section,  r  —  radius  of  gyration  of  the  section  about  its 
neutral  axis,  h  =  height  of  section. 


252  MECHANICS  OF   THE    GIRDER. 

By  equating  the  last  member  of  (332)  to  the  known  moment 
due  the  external  forces  applied  to  the  girder,  any  one  of  the 
four  quantities  Biy  S,  r,  ht  may  be  found.  But,  when  the  girder 
is  fixed  at  one  or  both  ends,  we  need  to  know  the  point  or 
points  of  contrary  flexure,  in  order  to  determine  the  end 
moments. 

101.  For  the  girder  of  uniform  height  and  strength,  fixed  at 
both  ends,  it  follows  from  the  uniformity  of  the  unit  strain  and 
height,  which  causes  a  uniformity  of  curvature,  that,  as  already 
stated,    each   point    of    contrary   flexure   is    sensibly   midway 
between  the  centre  of  gravity  of  the  load  and  the  correspond- 
ing end  of  the  girder. 

Assuming  that  the  height  and  strength  are  uniform,  and 
that,  for  any  required  form  of  cross-section,  the  necessary 
variation  in  its  area  is  attained  by  varying  the  thickness  of 
the  beam  only,  we  shall  have,  in  (332),  r,  k,  and  B,  constant, 
so  that  the  variable  area,  S,  may  be  found  at  once  for  any 
section  of  the  beam  ;  and  from  5  the  thickness  is  to  be 
determined. 

102.  Beam    of    Uniform    Strength   and   Height  fixed   at 
Both  Ends,  and  bearing  a  Concentrated  "Weight,  IV,  at  the 
Distance  a'  from  the  Left  End.  —  The  moment  at  any  point 
between  the  weight  and  left  end  of  the  beam,  that  is,  when 
x  is  not  greater  than  a'y  is  given  by  equations  (40),  (93),  and 
(332),  thus, 


(333) 


Now,  Mx  =  o  when  x  =  \a'y 

...    o  =  Wl  ~7a  a'  -  -(M,  -  M2)  +  Mt.       (334) 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  253 

Also,  when  x  is  not  less  than  a',  we  have,  from  (43),  (93), 
and  (332), 


.      (335) 


Mx  =  o  when  x  =  \(l  +  ^0> 
o  =  Wl-=-^a'  -  \(M,  -  M2)  -  ^(M,  -  M2)  +  M,.     (336) 

From  (334)  and  (336),  we  find 

a>).  (337) 


That  is,  the  end  moments  are  equal  and  negative  for  any 
given  position  of  the  load. 

Eliminating  Ml  and  Mt  from  (333)  and  (335),  we  obtain 


=  ^f-,      (338) 

=  mL^l(x  _  V).  (339) 

2    'h  **,  (340) 

(34i) 


254  MECHANICS  OF   THE   GIRDER. 

EXAMPLE  i.  —  If  the  varying  cross-section  is  a  rectangle  of 
the  breadth  b,  and  constant  height  //,  we  have  r2  =  y^2,  and 
(339)  and  (341)  become 


bh  =  -        (x  -  X)>  (342) 


+ 


a'         \ 

~  x}         (345) 


If,  further,  the  weight,  W,  is  at  the  centre  of  the  girder, 
a'  =    /,  and  when 

-  I/).  (346) 

~X)-  (347) 


i  WJ 
In  (346),  for  x  =  o,  b  =  b,  =  -  ^_^-,  at  left  end. 


x  =  £/,  ^  =  o,  at  quarter  point. 

#  =  i/,          ^  =  1^?-,  at  centre. 
^BJi2 

~  M7/ 

In  (347),  for  x  =  J/,  ^  =  •2-—,  at  centre. 


x  =  |/,  b  —  o,  at  quarter  point. 

-  ;  at  right  end. 


DEFLECTION.   END   MOMENTS,   ETC.,   FOUND. 


255 


If  the  beam  is  of  oak,  and  Bt  =  ^B  =  1,060  pounds,  E  = 
2,150,000,  /  =  i  So  inches,  h  =  15  inches,  W  =  4,000  pounds, 

3  X  4000  X  1 80 
tb^A=r^=     -*,=     ~4  x  I06oX  IS2"    -2-264  inches; 

the  algebraic  sign  only  indicating  the  direction  of  the  inclina- 
tion of  the  vertical  planes  forming  the  sides,  to  the  vertical 
longitudinal  plane  of  the  r3£am. 

Fig.  99  shows  this  beam  thus  loaded,  in  plan  and  elevation. 

It  is  evident  that  the  deflection  of  the  part  BD  =  |7,  or  of 
the  part  AB  =  \l,  as  a  semi-beam,  is  equal  to  the  deflection 
of  a  beam  of  uniform  strength  and  height  supported  but  not 
fixed  at  the  points  B  and  D,  and  bearing  the  concentrated 
weight  W.  But,  by  equation  (307),  the  deflection  of  the  part 
AB  or  BD  is,  since  for  x  we  must  put  \l, 


D 


FIG.  99. 
Therefore  the  total  deflection  at  Cy  the  centre  of  AEt  is 


equal  to  -  IO  °   —  —  -  =  0.1331  inch  in  the  present  case. 
8  X  2150000  X  15 


256  MECHANICS  OF  THE   GIRDER. 

At  the  points  of  contrary  flexure,  where  b  =  o,  the  beam, 
of  course,  must  be  enlarged,  to  resist  with  safety  the  shearing- 
strains. 

The  shearing-strain  at  each  of  these  points  is  now  J  W  = 
2,000  pounds.  By  Table  I.,  article  42,  the  ultimate  shearing- 
strength  of  oak  across  the  grain  is  4,000  pounds  to  the  inch  ; 
or  the  working-strength  is  400  potfnds  to  the  square  inch  of 
cross-section. 

We  require,  therefore,  at  least  %£$£•  —  5  square  inches  of 
area  at  each  point  of  contrary  flexure  ;  that  is,  the  beam,  being 
15  inches  deep,  must  be  at  least  J  inch  thick  at  these  points, 
even  when  restrained  from  moving  laterally. 

103.  Beam  of  Uniform  Strength,  Height,  and  Load,  fixed 
Horizontally  at  Both  Ends.  Rectangular  Cross-Section. 

Equations  (49)  and  (93)  give 


Mx  =  >(/  -  *)*  -      l  ~      Z*  +  Ml  =  \BJh\    (349) 


Make  x  =  o,  then 

M  =  M2  =  \BJ>Ji\ 

But,  as  in  article  102,  b  —  o  when  x  =  \l  or  f  /, 
...    %w(i  _  i/)i/  +  M,  =  o, 
Mt  =  M2=  --&0//2, 

Mc  =  \w(l  -  V)V  ~  -h™1*  =  ?W2> 

Mx  =  \w(l  -  oc)x  -  &wl*  =  \B,bh\  (350) 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  2$? 


When  x  =  o,  or  x  =  /,  (351)  gives 


which  is  the  width  of  beam  at  either  fixed  end  ; 


which  is  the  width  of  the  beam  at  centre,  being  one-third  of  the 
width  at  either  end. 

Equation  (351)  being  that  of  a  parabola  with  respect  to  the 
variables  b  and  x,  the  plan  of  this  uniformly  loaded  beam,  of 
uniform  strength  and  height,  fixed  at  its  ends,  is  shown  in 
Fig.  100. 

I 


ELEVATION. 


PLAN. 


FIG.    100. 

EXAMPLE.  —  Oak  beam.  /  =  180  inches,  h  =  15  inches, 
ends  fixed. 

Take  Bl  =  1,060,  E  =  2,150,000,  w  =  &$$•  =  44$  pounds 
to  the  linear  inch. 

Then 


j.  =  M2  =  —  _  x          x  i8o2  =  —135000  inch-pounds. 
32         9 


Mc  =  45000  inch-pounds. 


258  MECHANICS  OF   THE   GIRDER. 

b,  =  b2  =   -  9  *  -F  *   i8o2   =  _  .n>  =  thicknesg  at 

16  X  1060  x  is2 

bc  =  1.1321  inches,  at  centre. 

The  maximum  deflection  is  evidently  given,  as  in  article 
102,  by  equation  (348),  and  is 

D  =  0.1331  inch. 

104.  Concentrated  Weight,  W,  at  the  distance  a'  from 
the  Unfixed  End  of  a  Beam  of  Uniform  Strength,  fixed  at 
the  Right  End,  but  simply  supported  at  the  Left.  —  The 
point  of  contrary  flexure  must  be  at  the  distance  ;FO  =  l(/  -f-  cf) 
from  the  unfixed  end,  in  order  that  the  greatest  positive 
;moment  may  be  equal  to  the  greatest  negative  moment. 

Equations  (43)  and  (93)  apply,  giving,  since  M,  =  o, 


Mx  =  W-=-af  +  M     =  \BJ>h*  (352) 


.if  the  cross-section  be  rectangular,  and  x  'y  a'. 
For  x  =  /,  M2  =  \BJ)Ji\ 


.For  *  =  £(/  +  a'),    M2  =  -  Wa  '^  ~~  a     when  b  =  o. 

~ 


DEFLECTION,   END  MOMENTS,   ETC.,  FOUND. 


When***', 


=  *>%'('  ~  f  > 
^V    b+*  I 


Which  shows  that  the  width  at  Z>,  Fig.  101,  is  the  same  as  the 
width  at  C  for  h  constant. 

When  x  ^  a',  use  (40)  and  (93),  giving 

Mx  =  W*-=-£x  +  ^  =  i^M*.  (354) 

To  find  the  lowest  point,  E,  in  the  curve,  Fig.  101,  we 
equate  the  deflection,  D»  between  the  lowest  point  and  left  end 
of  the  beam,  to  the  total  deflection,  D2  +  Z?3,  between  the 
same  point  and  the  right  end  of  the  beam,  and  solve  the  equa- 
tion Z>x  =  A  +  Dy 

For  the  length  AE  =  z,  (307)  gives 


EB  =      l  +  a'   -  z 


BD  =         - 


JkLfi 


which  is  the  deflection  at  the  lowest  point,  E. 


26O  MECHANICS  OF  THE   GIRDER. 

EXAMPLE.  —  Given  W  =  4,000  pounds  at  the  distance 
a'  —  \l  from  the  unfixed  end,  A,  Fig.  101 ;  /  =  180  inches; 
h  =  15  inches  =  uniform  height  of  beam ;  B^  =.  -^B  =  1,060 
pounds  =  working  inch  strain  for  oak  ;  E  =  2,150,000.  Cross- 
section  rectangular. 

Then  width  of  beam  is, 

At  left  end,  (354), 

x  =  o,  b  =  o. 

At  the  weight,  (353), 

,      4000  x  1 80 

*  =  «  *  =   IQ6o  x  15-  =  3'°19 

(353), 

*  =  |/,  b  =  o. 

At  fixed  end,  (353), 

j  =  _  4000  X  180  =  _ 
1060  x  i52 

the  negative  sign  simply  showing  that  the  lines  cd,  Cjdu  have 
crossed  somewhere  between  x  =  \l  and  x  =  /. 
Moment  at  fixed  end, 

Jlf2    =    —  4Q003 ^-   =    —120000. 

1  +  ¥ 
Moment  at  the  weight, 

=  120000  inch-pounds. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  26 1 


The  deflection  at  the  lowest  point,  Ey  is  given  by  (355), 
1060  x 


~ 

Dl  = 


4  x  2150000  x  15  X  (|) 


=  0.1849 


C      E 


FIG.  101. 

105.  Continuous  Uniform  Load,  wly  on  a  Beam  of  Uni- 
form Strength,  fixed  at  the  Right  End,  and  simply  sup- 
ported at  the  Left,  Fig.  101.  —  The  figure  shows  the  curvature 
and  the  plan,  when  the  section  is  rectangular  ;  and  equations 
(49)  and  (93)  give,  since  MI  =  o, 


Mx  = 

for  these  conditions. 
For  x  =  /, 


(356) 


wl2 


For  x  = 


M2  =  - 


b  =  o. 


262  MECHANICS  OF  THE   GIRDER. 


For  *  =  M  =         *>/2  b  = 


b  =  f^(2l  -  Zx)x.  (357) 

Hence  the  breadth  at  £7  is  one-third  of  that  at  Z>  when  the 
height  is  uniform,  as  seen  in  the  parabolic  plan,  Fig.  101, 
derived  from  equation  (357). 

The  deflection  at  E  is  given,  by  (355),  for  this  case  also, 
provided  we  put  \l  for  a'. 


EXAMPLE.  —  Let  /  —  180  inches,  h  =  15,  B^  =  ^B  =  1,060 
pounds  =  working  unit  strain  for  bending  oak,  E  =  2,150,000. 
Cross-section  rectangular  ;  load  wl  =  8,000  pounds  uniformly 
distributed  continuously,  44$  pounds  to  the  inch. 

From  (356)  and  (357), 

x  =  o,  Ml  =  o. 

x  —  \l,  M  =  T^  X  ^  X  i8o2  =  80000  inch-pounds. 

x  —  |/,  M  =  o. 

#  =    /,  J/2  =  —  |  X  ^  X  i8o2  =  —240000  inch-pounds. 

Width  at  left  end, 

^  =  o, 

by  (357)- 

Width  at  i/, 


x 

=  «.oi  26  inches. 


3X  1060  x 

Width  at  |/, 

J  =  o. 
Width  at  right  end, 

b2  =  6.0378  inches. 


DEFLECTION,   END  MOMENTS,   ETC,,   FOUND.  263 

Put  -J/for  tf'in  (355),  and  find  the  deflection  D  =  0.1849  at 
lowest  point. 

106.  Beam  of  Uniform  Strength  and  Uniformly  Varying 
Height,  fixed  at  Both  Ends.  —  The  end  moments  Ml  =.  M2 
are  determined  for  this  case  as  in  articles  102  and  103,  for  the 
same  kind  of  load. 

To  find  the  deflection  of  this  beam,  we  may  regard  it  as 
composed  of  four  semi-beams,  Fig.  102. 

ist,  AJ3,  fixed  at  A  •  deflection  A- 

2d,  BC,  fixed  at  C,  the  lowest  point;  deflection  Z>2. 
3d,  CE,  fixed  at  C,  the  lowest  point ;  deflection  Dy 
4th,  EF,  fixed  at  F;  deflection  Z>4. 

Now  we  must  have  D*  +  D2  =  D3  +  D#  from  which  the 
lowest  point  and  its  deflection  may  be  found. 

EXAMPLE  i.  —  Take  one-half  of  the  girder  shown  in  Fig.  65, 
and  suppose  the  ends  of  this  half  to  be  immovably  fixed.  Call 
the  length  /  =  100  feet  =  1,200  inches ;  and  height  at  left  end, 
hQ  —  1 80  inches ;  height  at  right  end,  k,  —  300  inches.  Let 
the  girder  be  of  wrought-iron,  and,  as  in  article  93,  take  Bt 
=  i(£i  ~H  ^i)  —  9>ooo  pounds,  E  =  25,000,000 ;  and  suppose 
the  load  to  be  a  concentrated  weight,  W  =  200,000  pounds 
at  the  centre,  no  account  being  here  taken  of  the  girder's  own 
weight. 

By  article  100,  the  points  of  contrary  flexure  are  ^^  =  2$ 
feet  from  the  centre  of  the  beam ;  and  from  equation  (337), 
since  a'  —  £/, 

M,  =  M2  =  —  Mc  =  —  \  x  200000  x  £  X  \  X  1200, 
=  —30000000  inch-pounds. 


264  MECHANICS  OF  THE   GIRDER. 

The  area  5  of  any  cross-section  on  the  left  of  the  weight  is 
given  by  (339),  and  on  the  right  by  (341).  But  these  equations 
suppose  the  section  of  the  top  chord  to  be  equal  to  that  of  the 
bottom  chord  in  the  same  vertical  plane  of  section,  and  at 
the  centre  give 

Sc  =  200QOO  X  240  X  600  X  j  X  1200  =  inches 

2    X    QOOO    X    1200    X    J    X    2402 

at  left  end,  (339), 

£  =  200000  x  180  X  600  X  -j  X  1200  =  incheg 

2  x  9000  X  1200  X  J  X  i8o2 

at  right  end,  (341), 

S2  =  20000Q  X  300  X  600  x  I  X  1200  =  22>222  inches . 
2  X  9000  X   1200  X  J  X  3002 

the  negative  sign  indicating  only  a  difference  in  the  direction 
of  the  lateral  faces  of  the  chords,  that  is,  change  of  slope 
laterally. 

But  if  S'  =  area  of  section  of  chord  in  compression,  S"  = 
area  of  section  of  chord  in  tension,  we  have 


C,S'  =  P  =  --,  (358) 

cos  a 


TVS"  =  U  =  •£-*  (359) 

cos/5 

H       =  M  +  h, 
according  to  the  notation  and  equations  of  article  49. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  265 

From  which, 

S'  =  — — — ,  (160) 

Cih  cos  a 

S"=T^-  <36I) 

Calling  Ci  =  8,000  pounds,  7",  =  10,000  pounds,  a  being  the 
inclination  of  the  top  chord  for  all  parts  between  the  points  of 
contrary  flexure,  while  (3  =  o,  and  ft  being  the  slope  of  top 
chord  for  the  remainder  of  the  beam,  while  a  =.  o,  we  have,  at 
either  end, 

tan£  =  25IQQ15  =  o.i,     cos/3  =  0.99503 ; 

tana  =  o,  cos  a  =  i. 

At  centre, 

tana  =  o.i,     cos  a  =  0.99503; 

tan/3  =  o,        cos/3  =  i. 
At  left  end,  (361),  area  of  top  section, 


s,,  =  -  30000000  -  =  i 

10000  x  180  x  0.99503 

• 
At  left  end,  (360),  area  of  bottom  section, 


s,  =  = 

8000  X   180  X   i 


At  centre,  (360),  area  of  top  section, 

5,  =  -  30000000  -  =  I5         inches. 
8000  X   240  X  0.99503 


266  MECHANICS  OF   THE   GIRDER. 

At  centre,  (361),  area  of  bottom  section, 

30000000 
S"  =  loooo  x  240  X  i  =  I2'5°°  inches" 

At  right  end,  (361),  area  of  top  section, 

30000000 
=  10000  x  300  x  0.99503  =  Ia°5°  mches- 

At  right  end,  (360),  area  of  bottom  section, 

s,  =  -  3°oooooo        =        QO  inches_ 

8000  X   300  X    I 

The  totals  are  :  — 
At  left  end, 

S*  =  37-582  inches; 
at  centre, 

Sc  =  28.203  inches  ; 
at  right  end, 

S2  =  22.550  inches; 

differing  somewhat,  as  was  to  be  expected,  from  the  areas  com- 
puted on  the  supposition  of  equal  top  and  bottom  sections. 

The  deflection  for  each  part  of  this  girder  is  given  by  (311). 
See  Fig.  102. 

ist,   ABt  fixed  at  A  ;   /^  =  180   inches,  h  =.  h0  —  210, 

-i  —  -,  7z0  —  hi  =  30,  and  we  have 
h       7 


2    X    QOOO    X    3002 


(  /  7 

{  I8°  -  2IO(I  -  ^302585  log  g 


25000000  x  30* 

=  125  x  2'37IS  =  0'I?I  inch- 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  267 

4th,    EFy    fixed   at   F\     //,  =  300,  h  —  hQ  =  270, 

j  hi        10 

A,  -  //,  -  -30,  -^  =  -, 

2X  9000x300'  •   I         _  270/2.302585  IogI9  +  ,\\ 
25000000  x  (-3°H  V  9          /) 

o 
=  -y-  X   1.5528  =  o.i  12  inch. 

Now,  in  equation  (311),  -  -  —  is  constant,  since  the  length, 

//o  —  //! 

/,  varies  as  the  height,  h.     Therefore,  — 

* 

2d,  BC,  fixed  at  C\  h  —  210, 


y  =  A  =  yf-O^  -  210  x  2.302585  log  ^ 

-f    210    X    2.302585  log  210   —    210). 

3d,  CE,  fixed  at  C  ;  /*  =  270, 


2.302585  log  A, 

H-  270  x  2.302585  log  270  -  270), 
and 

A  +  A  =  A  +  A- 

After  making  the  substitutions,  and  reducing,  we  find 
hi  =  236.15  inches, 

which  is  the  depth  of  the  girder  at  lowest  point,  C.    Hence  dis- 
tance of  lowest  point  from  left  end  is 

10(236.15  —  180)  =  561.5  inches. 

A  =  o.i  08  inch,     £>3  =  0.167  inch. 

A  +  A  =  A  +  A  =  0.279  inch  at  C 


268  MECHANICS  OF  THE   GIRDER. 

EXAMPLE  2.  —  Take  the  same  girder  as  in  the  preceding 
example,  but  let  the  load,  W  =  200,000  pounds,  be  75  feet  from 
the  left  end.  Then  the  points  of  contrary  flexure  are,  where 
x  =  |/,  and  #  =  •!/,  and  by  (337),  since  now  a'  =  |/, 

MI  —  M2  —  —  \  X  200000  X  I  X  J  X  1200 
=  —22500000  inch-pounds. 

At  the  weight,  x  =  a'  =  f /,  (338)  gives 

M  =  200000  X  J  X  f  X  1200  =  22500000  inch-pounds. 
At  the  centre,  x  =  |7, 

Mc  =  200000  x  \  X  \  X  1200  =  7500000  inch-pounds. 
At  left  end,  by  (361),  area  of  top  section, 

S»  = 22500000 =  12.56  inches. 

10000  X  180  X  0.99503 

At  left  end,  by  (360),  area  of  bottom  section, 

S'  =  225°°000         =  15.63  inches. 

8000  x  180  X  i 

At  the  weight,  (360),  area  of  top  section, 

S'  =  2250000° =  10.47  inches. 

8000  X   270  X  0.99503 

At  the  weight,  (361),  area  of  bottom  section, 

S"  =  225°0000 =  8.33  inches. 

10000  x  270  X   i 


DEFLECTION,  END  MOMENTS,  ETC.,  FOUND.  269 

At  right  end,  (361),  area  of  top  section, 

221500000 
S"  =  10000  x  300  x  0.99503  =  7-54  mches. 

At  right  end,  (360),  area  of  bottom  section, 

s,  =  22500000 =       g  inches_ 

8000  X   300  X   I 

Since  equations  (338)  and  (340)  are  of  the  first  degree  with 
respect  to  M  and  x,  and  since  h  varies  uniformly  with  x,  we 


PLAN  OF  TOP  CHORD. 


PLAN  OF  BOTTOM  CHORD. 
FlG.  102. 


have  M  and  5  in  (360)  and  (361),  varying  uniformly  from  the 
ends  of  the  girder  and  from  the  weight  to  the  points  of  con- 
trary flexure,  as  shown  in  plan  of  chords,  Fig.  102,  where  the 


270  MECHANICS  OF  THE   GIRDER. 

depth  of  each  chord  is  supposed  to  be  uniform,  and  the  varia- 
tion in  size  of  chord  attained  by  varying  the  thickness  only. 

The  deflection  for  each  of  the  four  semi-beams  into  which 
the  girder  now  becomes  divided,  is  to  be  found  as  in  the  preced- 
ing example,  where  the  load  was  at  the  centre. 

We  now  have,  for  the  part  AB  =  f/,  fixed  at  A,  /it  =  180, 
h  =  hQ  —  22$,  k,  -j-  k0  —  f,  k0  —  /i,  =  45  ;  and  (311)  becomes 


2  x  9000  x 
y  ==  Dr  = 


25000000  x 


45o2  j  /  5\) 

-^—  \  1  80  -  225    i  -  2.302585  log-    > 
452   (  °\  4/) 


=  -  X  5.206  =  0.37486  inch. 


For   the   fourth   part,   EF  —  \l,   fixed   at  F]    7/x  =   300, 
h  =  h0  =  285,      h,  -f-  h0  =  f^,      //o  —  ?h  —  —15  ; 


y  =  A  =  iFs^00  -  285T  +  2.302585  log         =        x  0.382 

=  0.0275  inch. 

For  the  second  part,  BCy  h  =  h0  =  225, 


A  =  -dhK7*'  -  225  x  2.302585  log^x 

+  225  x  2.302585^225  -  225). 

For  the  third  part,  CE,  h  —  k0  =  285, 


y  =  A  =  T§*(*«  -  285  X  2.302585  log  Ax 

H-  285  x  2.302585^285  -  285). 

And  since  A  +  D2  =  Dz  +  £>4,  we  find 

h^  =  234.761  inches; 
that    is,    the     lowest     point,    C,    is     now     at     the    distance 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  2/1 

10(234.761  —  1 80)  =  547.61  inches  from  the  left  end  of  the 
beam.  Using  this  value  of  hu  we  find  D2  and  Dy  and  have 
finally, 

A  =  o-3749>    A  =  0-3622, 
D2  =  0.0148,    D4  =  0.0275, 


Deflection  at  C    =  0.3897  inch  =  0.3897  inch; 

an  apparently  paradoxical  result,  since,  when  the  same  load, 
W  =  200,006  pounds,  was  at  the  centre  of  the  girder  of  uni- 
form strength,  and  having  the  same  varying  height  and  same 
length,  /  =  100  feet,  the  deflection  was  only  0.279  inch  at  the 
lowest  point.  The  paradox  vanishes,  however,  when  we  take 
into  account  the  difference  in  the  length  of  the  component 
semi-beams  for  the  two  cases.  Indeed,  it  may  be  easily  shown 
that  a  girder  of  uniform  height  and  strength,  bearing  a  concen- 
trated load,  both  ends  being  fixed,  deflects  least  when  that  load 
is  at  the  centre,  and  the  four  component  semi-beams  are  of 
equal  length. 

Suppose  that,  in  (307),  we  have,  for  — 

First  semi-beam, 

x  =  \a',  according  to  article  100; 
second  semi-beam, 


third  semi-beam, 


fourth  semi-beam, 


2/2  MECHANICS  OF  THE   GIRDER. 

Then,  if  u  is  half  the  sum  of  the  four  deflections,  that  is,  if 
u  =  the  total  deflection  of  the  beam,  we  have 


+  W  -  a  )2].      (362) 

XJJsfl, 

Put 


Therefore  a'  —  \l  renders  u  a  minimum,  since  2ar  is  positive 
and  /  constant. 

In  a  similar  manner,  from  (311),  may  the  position  of  the  load 
be  found  on  the  beam  of  uniform  strength  and  uniformly  vary- 
ing height,  the  ends  being  fixed,  when  it  is  required  to  know 
what  position  of  a  given  load  gives  the  least  deflection. 

EXAMPLE  3.  —  Continuous  uniform  load  wl  =  400,000 
pounds  upon  the  same  girder,  Fig.  102.  Since  the  moments 
of  the  external  forces  are  independent  of  the  height,  equation 
(350)  applies  here,  giving  for 

x  =  o,     M!  =  —  -/$  X  400000  x  1200  =  —45000000  inch-pounds; 
x  =  \lt    Mc  =  J-  X  400000  x  1200  +  M!  =  15000000  inch-pounds; 
x  =    /,    M2  =  MI. 

By  equations  (360)  and  (361),  we  find  — 

At  left  end,  section  of  top  chord, 

415000000 
S   =10000  x  180x0.99503  =  2S'I2S  mches' 

At  left  end,  section  of  bottom  chord, 

=  3I.2SO  inches. 
8000  X  180  X  I 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  2/3 


At  centre,  section  of  top  chord, 

S'  =  __  15000000  -  =  7.8S1  inches. 
8000  x  240  x  0-99503 

At  centre,  section  of  bottom  chord, 

15000000 
S    =  xoooo  x  240  X   i  =  6'25°  mches- 

At  right  end,  section  of  top  chord, 


45000000 
S   =  zoooo  x  300  X  0.99503  =  '5-075 


At  right  end,  section  of  bottom  chord, 

45000000 
5  =  8000  x  300  x  i  =  I 


inches- 


The  deflection  must  be  the  same  as  in  example  i  ;  viz., 
Dl  +  D2  —  0.279  inch,  since  the  centre  of  gravity  of  each  of 
the  two  loads  is  at  the  same  point,  and  the  unit  strain  the 
same. 

107.  Beam  of  Uniform  Strength  and  Uniformly  Varying 
Height,  fixed  at  One  End,  and  simply  supported  at  the 
Other.  —  Since  the  position  of  the  point  of  contrary  flexure 
depends  upon  the  moments  due  the  external  forces,  which 
moments  are  independent  of  the  height  of  the  girder,  we 
already  have,  in  articles  104  and  105,  the  point  of  contrary 
flexure,  and  the  moment  at  the  fixed  end,  M2,  for  the  present 
case  of  uniformly  varying  height,  if  the  load  be  either  concen- 
trated or  uniform  and  continuous. 

The  cross-section  at  any  point  is  given  generally  by  equation 
(332),  the  deflection  of  each  of  the  three  component  semi-beams 


274  MECHANICS  OF   THE   GIRDER. 

by   (311),  and  the  equation  Dl  =  D2  +  D3  fixes  the  lowest 
point. 

EXAMPLE  i. — Take  a  girder  of  the  same  varying  height 
and  same  length  .as  in  the  examples  of  article  106,  Fig.  102,  of 
wrought-iron,  but  now  fixed  at  the  right  end  and  simply  sup- 
ported at  the  left ;  that  is,  let  E  =  25,000,000,  B^  =  9,000, 
Ct  =  8,000,  Tt  —  10,000,  /  =  1,200  inches,  hQ  =  180  inches 
=  height  at  left  end,  //,  =  300  inches  =  height  at  fixed  end, 

tana  = =  o. i  =  tan  of  slope  of  top  chord,  cos  a  = 

0.99503,  tan/3  =  o,  cos (3  —  I,  since  bottom  chord  is  horizontal. 
Let  the   load    W  =   200,000  pounds   be   at   the   distance 
a'  =  |-/  from  the  unfixed  end,  the  point  of  contrary  flexure 
.being  at  x  =  £(/  +  af)  =  \L     Then,  from  (352), 

Moment  at  fixed  end  =  M2    =  —  200000- | 

=  —32000000  inch-pounds. 

Moment  at  weight,  Maf  =       32000000  inch-pounds. 

Moment  at  left  end,        MI    =      o. 

At  the  left  end,  (360)  and  (361)  give  the  chord  cross-sec- 
tions =  o ;  but,  of  course,  as  before  shown  and  exemplified  for 
all  such  cases,  the  end  must  be  enlarged  to  bear  the  shearing 
and  crushing  strains  with  permanent  safety. 

At  the  load,  (360)  gives 

S'  =  15.46  inches  =  section  at  top. 
At  the  load,  (361)  gives 

S"  =  12.31  inches  =  section  at  bottom. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  2?$ 

At  fixed  end,  (361)  gives 

S"  =  10.72  inches  =  section  at  top. 
At  fixed  end,  (360)  gives 

S'  =  13.33  inches  =  section  at  bottom. 

Applying  equation  (311)  to  the  three  parts  of  this  beam, 
AB,  BD,  DE,  we  find  the  deflection,  Fig.  103,  — 

AB,  fixed  at  B\  JiQ—  180, 
y  =  D,  =  TM/^  -  180  x  2.302585  (log^,  -  log  1  80)  -  180]. 

BD,  fixed  at  B\  hQ  —  280, 
y  =  D2  =  -rf^/*!  —  280  x  2.302585  (log  hi  —  log  280)  —  280]. 

DE,  fixed   at  E\   hQ  =  280,   //,  =  300,  7/0  —  //,  =  —20, 

&  =  IS 

"  14 


y  =  Dz  =  yMsoo  -  280(2.302585  log  if  +  i)]  =  yfj  X  0.681 
=  0.049  inch. 

From  the  equation  D1  =  Z72  +  Z)3  we  have 

^!  =  229.731  inches, 
/.     DI  =  0.419  inch  =  deflection  at  lowest  point,  B. 

EXAMPLE  2.  —  Take  the  same  girder,  with  the  same  condi- 
tions, as  in  example  i,  except  that  the  load  is  now  wl  =  400,000 
pounds,  uniformly  distributed. 


276 


MECHANICS  OF  THE   GIRDER. 


The  moments  are  found  by  (356) ;  thus, 
x  =  o,  M  =  MI  =  o. 

17  n/r    400000  X  1200     /:/-/-/-/-£/: 

X  =  -|/,  M  =  "   =  26666666. 

Io 

x  =  /,    J^  =  —80000000  inch-pounds. 


PLAN,  UNIFORM  LOAD. 
FIG.  103. 

Using  these  moments,  we  find  the  required  cross-sections, 
by  means  of  equations  (360)  and  (361),  as  follows  : 

x  =  %l,  section  of  top  chord,  S'  —  15.23  square  inches. 
x  =  |/,  section  of  bottom  chord,  S"  =  12.12  square  inches. 
x  =  /,  section  of  top  chord,  S"  =  26.80  square  inches. 
x  =  /,  section  of  bottom  chord,  S'  —  33.33  square  inches. 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  2JJ 

Deflection  for  the  three  parts,  by  (311) :  — 

FG,  fixed  at  G ;  //0  =  1 80  inches, 
y  =  D,  =  TtbC/Zx  -  1 80  x  2.302585 (log ^  -  logiSo)  —  180]. 

GH,  fixed  at  G ;  /i0  =  260  inches, 
y  =  Z>2  =  yf^-^1  —  260  X  2.302585  (log hi  —  log  260)  —  260]. 

HI,  fixed   at  /;   hQ  =  260,  //,  =  300,   h0  —  /*,  =  —40, 

&  =  iS 

4,    13' 

^  =  A  =  rMs00  -  260(2.302585  log ^f  +  i)]. 

From  A  =  A  +  Z>3,  we  find  /zx  =  226.547  at  the  point 
x  =  10(226.547  —  1 80)  =  465.47  inches, 

.*.    Z^,  =  0.37  inch 
at  G,  Fig.  103. 

SECTION  5. 
Camber. 

108.  Camber  is  the  slight  upward  curving  or  crowning  that 
is  sometimes  given  to  a  girder,  in  order  to  obviate  the  sagging 
which  would  otherwise  result  from  the  deflection  of  the  same 
girder  made  without  this  slight  arching.  The  effect  of  camber 
is,  therefore,  to  keep  the  track  line  straight  under  the  working- 
load,  and  thereby  prevent  that  increase  of  stress  which  would 
otherwise  be  developed  by  the  falling  and  rising  of  loads  mov- 
ing rapidly  along  a  line  originally  straight.  In  no  other  respect 


2/8  MECHANICS  OF  THE   GIRDER. 

does  camber  augment  the  efficiency  of  the  structure.  Some- 
times, however,  a  greater  upward  curvature  than  that  here  con- 
templated is  given  to  the  floor  line  of  highway  bridges,  as  being 
more  pleasing  to  the  eye ;  but  so  large  a  convexity,  if  effected 
in  the  girder  itself,  is  always  at  the  expense  of  material  or  of 
efficiency,  as  will  appear  from  a  comparison  of  the  capabilities 
of  two  girders  shaped  like  Figs.  23  and  80,  of  equal  length  and 
equal  height  between  axes  of  chords. 

It  is  evident  that  camber  may  be  given  to  the  floor  or  track 
line  in  three  ways :  — 

ist,  The  girders  may  be  made  in  normal  shape,  and  the 
floor  or  track  line  be  raised  sufficiently  to  counteract  the  deflec- 
tion due  the  total  load.  In  this  case  the  two  chords  of  each 
girder  will  sag,  while  the  cambered  floor  line  becomes  straight 
under  load. 

2d,  The  chord  which  carries  the  floor  line  may  be  cambered, 
while  the  other  is  built  in  normal  shape.  In  this  case  the 
uncambered  chord  will  sag,  while  the  other  assumes  its  normal 
shape  under  load. 

3d,  The  girder  may  be  so  built,  that,  before  the  load  is 
imposed,  its  proper  floor  line  will  have  a  deflection  equal  and 
opposite  to  the  deflection  due  the  total  load,  and  that  the  whole 
girder  will  assume  its  normal  shape  under  load. 

We  need  examine  and  exemplify  only  the  second  and  third 
cases. 

109.  Change  of  Length  calculated  from  the  Unit  Strain. 

—  If  Az  =  total  contraction  for  the  original  length  /„  and 
X2  =  total  elongation  for  the  original  length  /2,  of  any  strained 
member,  we  have,  within  the  elastic  limit  where  the  amount  of 
displacement  per  unit  of  length  varies  as  the  stress, — 


DEFLECTION,   END  MOMENTS,   ETC.,   FOUND.  279 

For  compressed  member, 

XV  T  /T  X  X  \ 

i  =  -jr j  (364) 

for  extended  member, 

^  =  ^r2;  (365) 

£/x  and  7^  being  the  allowed  unit  strains,  and  Ee  and  ^  the 
moduli  of  compressive  and  of  tensile  elasticity  respectively. 

The  total  difference  between  the  lengths  of  the  two  chords 
of  a  girder  after  deflection  is,  therefore, 

A  =  \v  +  A,  =  (%  +  fiV  (366) 


provided  the  chords  were  of  equal  length,  /,  before  deflection, 
and  of  uniform  strength. 

If  an  originally  straight  girder  of  equal  and  parallel  chords 
take  the  circular  form,  Fig.  104,  after  deflection,  the  neutral 
line  being  midway  between  the  chords,  we  must  have  for  it, 


...    X  =  X,  +  A2  =         +         =  (368) 

Ec        Et          E 

if  E  =  \(EC  -J-  E})  =  modulus  of  transverse  elasticity,  and 
BI  =  ^(Cl  +  71,)  =  bending  unit  strain. 

no.  Elongation  and  Contraction  calculated  from  Deflec- 
tion. —  Let  ABCD,  Fig.  104,  represent  an  open-built  semi- 
girder  fixed  horizontally  at  A  and  C,  and  having  its  deflection, 
D  =  NH,  greatly  exaggerated  in  the  figure  ;  the  actual  lines 
NFM  and  HM  being  sensibly  equal  each  to  /,  the  original 


280 


MECHANICS  OF  THE   GIRDER. 


length  of   the  parallel  chords  AB  and  CD  of   the  semigirder 
whose  height  is  AC  =  h. 


FIG.  104. 

We  may  without  appreciable  error,  for  the  present  purpose, 
regard  the  deflection  curve  as  circular. 

Take,  as  radii  of  curvature,  r  for  neutral  line  NM,  r  +  \h 
for  the  extended  chord,  r  —  J/z  for  the  contracted  chord. 
Then,  from  the  geometry  of  the  figure,  we  have 

/*  =  D(2r  -  D), 


r   = 


(369) 


since  D  is  very  small  compared  with  r. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  28  1 

Also,  from  the  figure, 

r  / 

r  +  \h       I  +  A/ 


(37o) 


which  is  the  length  to  be  given  to  the  chord  in  compression, 
the  figure  being  inverted.     Again, 


r  -\h       I  -  A,' 
...    /-A,    =/(y;^,  (37-) 

the  length  required  for  the  chord  in  tension,  the  figure  being 
inverted. 

Subtracting  (371)  from  (370),  we  find  the  total  difference  in 
length  required, 

X  =  X,  +  X2  =  ^=^,  (37*) 

after  eliminating  r  by  means  of  (369). 

B  I2 

It  may  be  noted  that   (368)  and   (372)  give  us  D  =  -~ 

k,h , 

which  is  equation  (308)  for  the  semi-girder  of  the  length  /. 

In  (369)  and  (372),  we  must,  of  course,  put  J/  for  /,  when  we 
apply  these  equations  to  a  girder  of  the  length  /  supported  at 
both  ends. 

in.  Suppose,  now,  that  it  is  required  to  camber  only  the 
straight,  horizontal  bottom  chord,  to  which  the  moving-load  is 
to  be  applied.  This  supposition  includes  Classes  II.,  IV.,  VII., 
IX.,  X.,  and  XII.  of  article  49. 


282 


MECHANICS  OF   THE   GIRDER. 


We  may,  by  the  formula  proper  for  the  given  girder,  find 
the  deflection  at  each  panel  point,  or  apex,  of  the  bottom  chord. 
If  we  now  assume  that  no  apices  of  the  bottom  chord  are  to  be 
moved  horizontally,  by  reason  of  the  adopted  camber,  we  must 
theoretically  increase  each  normal  panel  length,  c,  of  this  chord, 

in  the  ratio  — — = ;    8  being  the  inclination  to  the 

c  cos/3 

horizon  of  any  panel  length,  c  +  A£  of  the  bottom  chord  when 
cambered,  and  A£  being  the  change  of  length  in  the  bottom 
chord  for  any  panel,  by  reason  of  the  adopted  camber.  Also, 

tan  p  —  -,  D!  and  D  being  the  deflections  at  any  two 

consecutive  apices. 


FIG.  105. 

Then  each  vertical  member,  as  FB,  Fig.  105,  coming  down 
to  a  lower  apex,  B,  must  be  shortened  by  the  amount  of  deflec- 
tion, Dlt  computed  for  the  apex ;  and  each  diagonal  or  brace, 
BE  —  v,  terminating  at  the  same  apex,  must  be  shortened  in 
the  ratio 


-  =  l 

v       \ 


+  h2 


(373) 


where  h  is  the  normal  difference  of  level  between  the  ends  of 
any  diagonal. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  283 

In  this  case  the  effective  depth  of  the  girder  at  the  centre 
has  been  lessened  by  D. 

EXAMPLE.  —  Let  us  find  the  changes  of  length  required  to 
effect  camber  in  the  bottom  chord  alone  of  a  wrought-iron  para- 
bolic bowstring,  where  /  =  2,400  inches,  h  =  225,  n  —  16 
panels  of  c  —  ^§^  =  150  inches  each,  C,  =  6,698  pounds, 
T,  =  10,000  pounds,  B*  =  8,349  pounds,  and  E  —  24,000,000 
after  the  frame  has  taken  its  permanent  "  set ; "  but,  as  ex- 
plained in  article  90,  we  will,  for  the  present  purpose,  take 
E  =  16,000,000,  to  provide  for  any  sagging  that  might  other- 
wise be  caused  by  the  first  full  load,  beyond  what  the  elasticity 
of  the  frame  can  recover. 

By  equation  (319),  putting  \l  =  1,200  for  /,  we  have  deflec- 
tion at  centre, 

1.386295  x  8349  x  1200* 
D  =  16000000  x  225       -  =  4.6297  inches. 

Now  we  may,  by  using  equation  (31 8),*  find  the  deflection  at 
each  panel  point ;  but  it  will  be  practically  accurate,  and  more 
simple,  to  regard  the  cambered  bottom  chord  as  a  parabola, 
having  the  central  height  D  =.  4.6297  inches,  and  then  find,  by 
equations  (136)  and  (137),  both  the  normal  heights,  /i,  and  the 
height  of  each  lower  apex  after  camber  is  effected.  Thus,  (136) 
now  becomes 

y  -  /i  _  -^-\  x  4*6297 
V        24oo2/ 

for  bottom  chord,  and 

y  —  [  i  —  .4£_  )  x  225 
V        24007 

for  top  chord ;  the  origin  being  at  the  middle  of  the  bottom 


284 


MECHANICS  OF   THE   GIRDER. 


chord  in  its  normal  shape.     From  these  equations  and  (373) 
we  compute  Dt  z/,  /,  h,  in  inches. 


X 

h 

D 

AZ? 

h-D 

v-*w, 

VHT*r 

i/ 

z' 

0 

225.00 

4-63 

—0.07 

220.37 

220.44 

216.85 

266.64 

263.67 

ISO 

221.48 

4.56 

—0.22 

216.92 

217.14 

206.38 

263.91 

258.14 

300 

210.94 

4-34 

—0.36 

206.60 

206.96 

189.09 

255.60 

241.36 

45° 

19343 

3.98 

-0.5I 

189.45 

189.96 

164.77 

242.04 

222.82 

600 

168.75 

3-47 

—  0.65 

165.28 

165.93 

I33-64 

223.68 

200.90 

750 

I37.II 

2.82 

-0.79 

134.29 

135-08 

95.62 

201.86 

177.88 

900 

98.44 

2.03 

-0-95 

96.41 

97.36 

5o-7i 

174.76 

158.34 

1050 

52-74 

1.08 

-1.  08 

51.66 

52-74 

— 

— 

— 

Theoretically,  the  end  panel  lengths  of  bottom  chord,  where 
the  inclination,  /?,  is  greatest,  would  become 

(i5o2  +  1.0852)5  =  150.0039  inches. 


But  this  is  practically  equal  to  the  normal  length,  150  inches  ; 
hence  we  will  not  change  the  panel  lengths  of  the  bottom 
chord. 

It  will  be  perceived  that  the  girder  thus  cambered  becomes 
the  parabolic  crescent. 

Instead  of  computing  dimensions  as  above,  it  is  evident  that 
the  elevation  may  be  drawn  accurately,  on  a  large  scale,  from 
the  central  deflection  Z>,  and  h  and  /;  and  then  all  desired 
lengths  can  be  taken  off  as  accurately  as  the  work  will  be  "  laid 
out "  in  the  shop.  The  camber  curve  may  always  be  drawn 
circular  for  an  originally  straight  chord. 

112.  Similarly,  if  we  would  camber  only  the  straight  upper 
horizontal  chord  of  Classes  III.,  IV.,  VIIL,  IX.,  XL,  XII.,  of 


DEFLECTION,  END  MOMENTS,  ETC.,   FOUND. 


285 


article  49,  without  moving  appreciably  the  upper  apices  hori- 
zontally, we  must  increase  the  normal  length  of  each  vertical 


FIG.  106. 


member  by  the  deflection  due  at  its  place,  and  the  normal 
length  of  each  diagonal  in  the  ratio 


See  Fig.  106,  where  the  efficient  depth  of  the  girder  has  been 
increased  at  the  centre  by  the  value  of  the  deflection  D. 

EXAMPLE.  —  Let  us  invert  the  uncambered  girder  of  article 
in,  and  effect  the  same  amount  of  camber,  D  =  4.6297  inches, 
in  the  straight  top  chord  alone.  We  have  the  same  values  of 
D  and  h  as  before,  and  readily  find  the  required  lengths  of  ver- 
ticals and  diagonals,  in  inches,  numbering  from  the  centre. 


(374) 


.      (375) 


286 


MECHANICS  OF   THE   GIRDER. 


X 

h 

D 

h  +  D 

V^r+x 

i/ 

W-^r 

z' 

O 

225.00 

4-63 

229.63 

229.56 

274.22 

226.11 

271.34 

J5° 

221.48 

4-56 

226.04 

225.82 

271.09 

215-50 

262.57* 

300 

210.94 

4-34 

215.28 

214.92 

262.08 

197.77 

248.22 

45° 

I9343 

3.98 

197.41 

196.90 

247-53 

17273 

228.77 

600 

168.75 

3-47 

172.22 

I7I-57 

227.90 

140.58 

205.58 

750 

i37-ii 

2.82 

139-93 

I39-I4 

204.59 

101.26 

180.98 

900 

98.44 

2.03 

100.47 

99-52 

180.01 

54-77 

159.69 

1050 

S2-74 

i.  08 

53-82 

— 

— 

— 

— 

In  like  manner  may  we  effect  camber  in  a  straight  chord  of 
any  one  of  the  classes  cited  in  this  and  the  preceding  article. 
And,  if  it  is  required  to  preserve  the  normal  height  between 
chords  after  camber,  we  must  change  both. 

113.  When  it  is  desired  that  the  effective  depth  of  the 
girder  be  not  altered  by  the  camber,  then  both  chords  must  be 
displaced  vertically  by  the  amount  of  the  deflection  at  the  sev- 
eral apices,  and  in  the  opposite  direction. 

In  articles  in  and  112  we  have  made  no  appreciable 
change  in  the  length  of  either  chord  by  reason  of  camber ; 
and,  of  course,  the  length  of  each  chord  will  be  changed  as  the 
load  takes  out  the  camber. 

Strictly,  regarding  camber  as  the  inverse  of  the  operation 
performed  by  deflection,  we  should  increase  the  normal  length 
of  the  compressed  chord  by  Ax,  equation  (364),  and  diminish 
that  of  the  stretched  chord  by  X,,  equation  (365) ;  but,  since 
the  whole  change  of  length  required  in  either  chord  is  very 
small  for  each  panel,  we  shall  distribute,  in  the  present  case, 
the  whole  difference,  X  —  Xx  +  A^,  of  length  due  to  deflection 
additively  among  the  panel  lengths  of  the  compressed  chord. 
Hence  camber  thus  produced  will  require  no  change  in  the 
normal  panel  lengths  of  the  chord  in  tension,  which,  under  the 
load,  will  resume  its  normal  line,  but  increased  in  length  by  A2. 


DEFLECTION,   END   MOMENTS,   ETC.,   FOUND.  287 

At  the  same  time,  the  compressed  chord  loses  A,  of  its  incre- 
ment, and  retains  X2. 

This  change  of  length  in  either  chord  which  rests  upon  the 
points  of  support,  may,  if  necessary,  be  provided  for  in  the  same 
manner  that  provision  is  made  for  the  effect  of  change  of  tem- 
perature. The  length  of  any  vertical  member  is  not  to  be 
altered  appreciably  for  camber  in  this  case,  since  the  vertical 
displacement  of  each  chord  is  assumed  to  be  the  same  for  any 
given  value  of  x ;  and  the  slight  change  in  their  length  caused 
by  the  spreading  of  the  verticals  to  fit  the  change  in  the  com- 
pressed chord,  is  hardly  measurable. 

But  the  length  of  any  diagonal  member  will  be  changed  as 
follows  :  — 

Let  ABCE,  Fig.  107,  represent  any  one  of  the  n  normal 
panels  of  a  girder,  and  A'B'C'E'  the  same  panel  when  cambered 


B' 


FIG.  107. 


by  adding  -  to  -  =1  c,  the  horizontal  projection  of  the  chord 
n       n 

AB  in  compression.  The  panel  points  in  both  chords  are  dis- 
placed vertically  by  the  deflections  Dr  —  CCr  ==•  BB',  and 
Dr  +  x  =  EEf  —  AAf,  appreciably ;  and  the  points  A  and  B 


288  MECHANICS  OF  THE   GIRDER. 


are  removed  horizontally  by  the  space  -.    Hence  practically  we 

have 

E'C'  =  EC, 


A'B'  = 

n 

A'E'  =  AE, 


'=  BC, 


E'F  =  -,     ^C'=- 
n  n 


/3  being  now  the  inclination  of  EC  or  E'C'  to  the  horizon. 

=  |/£  +  AY  +  /^.  _  ^tanjSVj4,  (376) 

-  (377) 


Instead  of  -,  we  may  evidently  employ  equations  (364)  and 
n 

(365)  in  finding  the  proper  increment  for  any  panel  length  of 
compressed  chord. 

EXAMPLE.  —  Take  BC  —  kr  =  240  inches,  AE  =  hr  +  x  —  200 

inches,  ^/^  =  -  =  c  =  150  inches,  -  =  ^±A  —  0.12  inch, 
«  «  ?2 

A  =  0.06  inch,  tan  ft  =  l-  X  24°  ~  2°°  -  -•     Then 
2;z  2  150  15 

^'^=  AB  +  0.12, 

0'=  [(i5o.o6)2-f  (240  -  150  x  A)2]*  =  266.304  =  if 
=  [(i5o.o6)2  +  (200  +  2o)2]5  inches. 


PILLARS.  289  v 


CHAPTER  VII. 

/ 

PILLARS. 

SECTION  i. 
Strength  of  Pillars,  by  Rational  Formula. 

114.  Under  the  general  term  pillars  we  shall  include  col- 
umns, posts,  struts,  props,  braces  in  compression,  and,  in  a 
word,  every  member  in  a  structure  whose  function  it  is  to  resist 
compressive  force  applied  at  its  end,  and,  in  general,  in  the  line 
of  the  longitudinal  axis  of  the  member. 

It  is  assumed  that  a  pillar  has  no  lateral  support  or  pressure 
applied  between  its  ends,  except  when,  owing  to  an  unavoidable 
existing  lateral  force  (as,  for  example,  the  weight  of  a  horizontal 
strut),  a  counter-force  is  applied  as  a  balance.  But  a  pillar  may 
have  its  ends  in  the  conditions  known  as  round,  hinged,  flat, 
imbedded,  fixed ;  the  two  ends  being  in  the  same  or  in  differ- 
ent conditions.  Pillars  may  be  long  or  short,  solid  or  hollow ; 
may  have  a  uniform  or  variable  cross-section  of  any  desired 
form. 

Long  pillars  yield  chiefly  by  bending  and  breaking  across  ; 
short  blocks  of  ordinary  building  material  yield  by  being 
crushed  without  bending,  properly  so  called.  At  what  exact 
ratio  of  length  to  diameter  bending  first  takes  place  in  a  given 
material,  is  not  at?  present  very  definitely  ascertained  ;  but  it  will 
be  safe  to  assume,  in  the  present  state  of  our  knowledge,  that 
bending  will  occur  when  this  ratio  is  as  low  as  three  for  such 


290  MECHANICS  OF   THE   GIRDER. 

material  as  wrought-iron.  Experiment  has  shown  what,  per- 
haps, we  might  have  inferred  from  a  stalk  of  wheat,  —  that 
material  is  saved  by  using  hollow  instead  of  solid  pillars  to 
.-support  a  given  load. 

115.  Pillars  of  Uniform  Cross-Section. 
Let  /  =  length  of  pillar, 
h  =  least  diameter, 

r  =  least  radius  of  gyration  of  cross-section, 
5  —  area  of  cross-section, 
D  —  greatest  deflection  of  pillar  ; 
;all  in  inches. 

E  =  modulus  of  transverse  elasticity, 

C  =  crushing-strength  of  standard  specimen  of  the  ma- 

terial, 
P  =  breaking-weight    applied  at  the  end  of   the  pillar 

and  in  the  line  of  its  axis  before  deflection, 
p 
Q  =  —  =  breaking-weight  per  square  inch  of  cross-section  ; 

vj 

;all  in  pounds  per  square  inch. 

./  =  Sr2  =  least  moment  of  inertia  (so  called)  of  cross-section. 
Mx  =  moment  of  forces  developed  in  any  normal  cross- 

section  by  the  given  load  P. 
Ml  =  the  end  moment  at  the  lower  end  when  that  end  is 

fixed. 
M2  =  the  end  moment  at  the  top  when  the  upper  end  is 

fixed. 

Suppose  the  pillar  vertical,  Figs.  108,  109,  no,  and  take 
the  origin  of  rectangular  co-ordinates  at  the  lowest  point  of  the 
pillar's  axis,  which  call  also  the  axis  of  x  ;  that  of  y  being  hori- 
zontal. 

Then,  from  equations  (15),  (93),  and  (187),  we  have  the 
moment  at  any  height,  x, 


It,  =  -EI       =      '  ~      >x  -MI  +  fy,      (378) 


PILLARS.  291 


wherein  no  account  is  taken  of  the  modified  condition  of  every 
cross-section  due  to  the  longitudinal  pressure,  Q,  per  unit. 

Now,  since  the  full  unit  strength  of  the  cross-section  of 
the  unloaded  pillar  is  C,  and  the  remaining  unit  strength  of  the 
loaded  pillar's  cross-section  is  (C  —  0,  it  follows  that  the 
expression  for  the  moment  of  the  internal  forces  developed  in 

any  cross-section  must  be  diminished  in  the  ratio  — 7^- 
We  then  have 

Mx  =  -P?-£  =  — ^ — -x  -  Mi  +  Py       (379) 
ctoc^  c 

if 

2  =  EI(C  -  0  =  Ef*(C  -  0 
PC  QC 

There  will  be  three  cases,  according  as  we  consider  neither, 
both,  or  one  only,  of  the  ends  fixed. 

CASE  I.  —  If  neither  end  can  produce  any  moment,  MI 
=  M2  =  o ;  and  we  have,  from  (379), 

e^  =  _y.  (38o) 

Multiplying  by  2dy, 

dvd 


Integrating  this  equation,  and  putting  a2  for  the  arbitrary 
constant  of  integration, 


from  which 


292 


MECHANICS  OF  THE   GIRDER. 


Integrating  again  between  the  limits,  for  xy  o  and  /;   and 
for  y,  o  and  o ; 


-        =    £«:r, 


where  n  may  be  any  whole  number  ;  but,  in 
order  that  P  may  have  the  least  value  it  can 
have,  consistent  with  the  bending  of  the  pillar 
necessarily  assumed  in  establishing  equations 
(378)  and  (379),  n  must  be  equal  to  unity.  (See 
Rankine's  "Applied  Mechanics,"  p.  352.) 


-  Q 


Cl2 


(381) 


FIG.  1 08. 


which  is  the  formula  for  pillars  with  rounded 
ends  that  can  generate  no  end  moments,  Fig.  108.  The  curved 
line  shows  the  deflected  axis. 

CASE  II.  —  If  both  ends  of  the  pillar  are  equally  fixed, 
Fig.  109,  so  that  the  elastic  curve  at  each  end,  after  flexure, 
has  for  its  tangent  the  original  undeflected  axis,  th'en,  in  equa- 
tion (379), 

M,  =  M2, 
whence 

*§.-*-*  (382) 

Multiplying  by  2cfy,  equation  (382)  becomes 


PILLARS. 


293 


Integrating,  we  find 


dx2 


=  2M,y  -  Py*  +  a, 


(383) 


where  a,  the  arbitrary  constant,  must  vanish,  since  -f-  =  o  when 

dx 

y  =  o.     Hence,  from  (383), 


Integrating  again,  with  the  condition  that 
y  =  o  when  x  —  o,  there  results,  after  can- 
cellin <Jp  from  the  denominators, 


TT       /  o  \ 
_.      (384) 


Also  we   have  y  =  o  when  x  =  ly  so  that 
(384)  becomes 


7T  "?7T      .      7T 

-    =    ^  --  1  --    =     27T, 

222 


K^^^%&%) 

FIG.  109. 

to  be  consistent  with  the  permanence  of  /  and  with  the  least 
positive  value  of  P.     Therefore 


C/2 


(385) 


which  is  the  formula  for  pillars  with  both  ends  equally  and 
fully  fixed. 


294 


MECHANICS  OF  THE   GIRDER. 


CASE   III.  —  When   only  one   end   of   the   pillar  is   fixed, 
Fig.  no,  and  the  other  end  can  cause  no  end  moment,  we 
have   (say)  MI  —  o,   and   derive,   from   equation 
(379), 


=  (f*  - 


(386) 


FIG.  no. 


an  equation  whose  second  member  cannot  be  inte- 
grated "  without  specific  connection  between  x 
and  y"  unless  there  is  such  a  relation  among  the 
quantities  which  compose  it  that  the  member  shall 
be  wholly  a  function  of  x.  (See  De  Morgan's 
"  Differential  and  Integral  Calculus,"  p.  208.)  Not 
knowing  the  specific  connection  between  x  and  y, 
nor  \^hat  relation  among  its  component  quantities 
there  may  be  to  render  the  second  member  a 
11  function  of  x  only,"  I  shall  here  assume  a  con- 
nection expressed  by  the  equation  of  a  parabola ; 
viz., 

y  =  x  +  A*2,  (387) 


where,  since  y  =  o  when  x  =  /,  b  =  —  - ;   and  shall  attempt 

only  an  approximate  solution  for  this  third  case. 

Putting  this  value  of  y  into  equation  (386),  it  becomes 


dx2 


mx  - 


Integrating,  with  the  condition  that  -^  =  o  when  x  =  /, 
since  the  upper  end  of  the  pillar  is  now  fixed, 

im(x2  -  /2)  -  i/B(*3  —  /3)-         (388) 
ax 


PILLARS.  295 


Integrating  again  between  the  limits  o  and  yt  o  and  xt 
...    p?y  =  \m(^    -  l*3\  -  \Pb(£    _  /sA     (389) 

But  y  —  o  when  x  =  7;  hence,  from  (389), 

m  =  J/W.  (390) 

If  in  (388)  we  make  -^  =  o,  we  may  find  the  value  of  x 

which  renders  y  —  D  a  maximum.  Dividing  by  ^r  —  /,  and, 
by  means  of  (390),  eliminating  m  and  Pb,  we  derive  from  (388), 
when  y  is  a  maximum, 

*  =  xVC1  ±  V/33)  =  0.421534 

since  the  negative  value  is  not  here  admissible. 

Taking  x  =  0.421537,  j/  =  ^  +  bx2,  and  ^  =  —  -,  we  find 
from  (389),  after  restoring  «2  and  m,  and  reducing, 


(39I) 


2  2.5  1  1  .Sr2 


which  is  an  approximate  formula  for  pillars  having  but  one  end 
fixed. 

The  nearness  and  sufficiency  of  this  approximation  will  be 
examined  in  article  120. 

It  should  be  observed  here  that  equations  (381),  (385),  and 
(391)  are  applicable  to  all  pillars  that  yield  by  bending,  of 
whatever  uniform  cross-section,  and  of  whatever  material  con- 
structed. Examples  of  the  application  of  these  three  equa- 
tions may  be  found  in  the  tables  of  article  121. 


296  MECHANICS  OF  THE   GIRDER. 

SECTION  2. 

Hodgkinson's  Empirical  Formula  for  the  Strength  of  Cast-Iron  and 

Timber  Pillars. 

116.  The  eminent  English  experimenter  Mr.  Eaton  Hodg- 
kinson  deduced,  from  his  experiments  upon  pillars  of  cast-iron 
and  pillars  of  timber,  formulae  which  have  found  place  in  all 
works  on  applied  mechanics. 

Using  the  notation  of  article  115,  and  taking  those  values  of 
the  constants  which  have  been  adopted  by  such  writers  as 
Rankine  and  Humber,  Mr.  Hodgkinson's  formulae  for  the  ulti- 
mate strength  of  cylindrical  cast-iron  pillars,  where  the  length 
of  each  is  not  less  than  thirty  times  the  diameter  if  the  ends 
are  flat,  and  not  less  than  fifteen  times  the  diameter  if  the 
ends  are  rounded,  become,  — 

Solid  cast-iron  pillars, 


hollow  cast-iron  pillars, 

p  =  A>!^r->  (393) 

k,  being  the  pillar's  internal  diameter,  and  A  "  representing  the 
strength  of  a  pillar  I  foot  long,  and  I  inch  in  diameter,  and 
being  a  constant  for  a  given  quality  of  iron,  but  ranging  in 
value,  for  different  irons,  from  75,000  to  112,000."  The  mean 
values  of  A  adopted  by  Professor  Rankine  are,  — 
Solid  pillars  with  rounded  ends, 

A  =  14.9  tons  =  33376  pounds; 
solid  pillars  with  flat  ends, 

A  =  44.16  tons  =  98918  pounds; 


PILLARS.  297 


hollow  pillars  with  rounded  ends, 

A  —  13  tons  =  29120  pounds; 
hollow  pillars  with  flat  ends, 

A  =  44.3  tons  =  99232  pounds. 

It  hence  results  experimentally  that  "fixing"  both  ends  of 
a  pillar,  Fig.  109,  enables  it  to  support  about  three  times  the 
load  which  would  break  it  were  the  ends  unfixed,  Fig.  108,  and 
incapable  of  developing  moment.  For  a  pillar  fixed  at  one  end 
and  rounded  at  the  other,  Fig.  no,  Mr.  Hodgkinson  found  the 
strength  to  be  a  mean  between  the  two  strengths  of  the  same 
pillar  when  both  ends  are  rounded  and  when  both  ends  are  flat. 
We  then  have,  for  cast-iron  pillars,  — 

Solid,  one  flat  and  one  round  end, 

A  =  66147  pounds; 
hollow,  one  flat  and  one  round  end, 

A  —  64176  pounds. 

When  the  length  is  less  than  30  or  15  times  the  diameter 
respectively,  Mr.  Hodgkinson  first  finds  P  by  equations  (392) 
and  (393),  and  then  corrects  P  by  means  of  this  supplementary 
formula ;  Pl  being  the  corrected  value  sought. 

_        PCS  CS 


(395) 


298  MECHANICS  OF  THE   GIRDER. 

which  is  an  empirical  equation  identical  in  form  with   (381), 
(385),  and  (391),  analytically  established. 

117.  The  Hodgkinson  formula  for  the  ultimate  resistance  of 
pillars  of  oak  and  of  red  pine  to  crushing  by  bending,  as  adopted 
by  Professor  Rankine,  "Applied  Mechanics,"  p.  365,  is,  with 
our  notation,  article  115, 

<2  =  |  =  s°o  cj,  (396) 

a  formula  to  be  used  only  when  Q  <  C,  the  crushing-strength 
of  the  material,  Table  II.,  article  60. 

Applications  of  the  Hodgkinson  formulae  are  given  in  tables 
of  article  121. 

SECTION  3. 

Gordon's  Empirical  Formula,  with  Rankings  Modification. 

p 

118.  We   have,  in  article    115,  Q  —  —  =  the  direct  unit 

o 

pressure  of  the  load  upon  every  cross-section  of  the  pillar. 

Now,  if  Bl  is  the  additional  unit  pressure  due  to  bending- 
moment  upon  those  fibres  where  the  bending-moment  is  great- 
est, and  if  f  denote  the  greatest  intensity  of  unit  pressure,  we 
have 

/  =  <2  +  ^.  (397) 

Regarding,  with  reference  to  the  central  moment,  a  loaded 
pillar  of  uniform  cross-section  as  in  the  condition  of  a  beam 
supported  at  both  ends,  and  carrying  the  central  weight 


W  =   -j—>  since  equations  (15),  (46),  and  (187)  give  us 

M  =  PD  =  \Wl  =        4  (398) 


PILLARS.  299 


we  find 

Wl  =  8^,7  =  4&g/P 

h  /2     ' 

from  (211); 

~  ~ULti 

T> 

From  which,  for  a  given  value  of  — :, 


But  (398)  gives 

Bi — -  (399) 

y          o/& 

if  /  =  kk2S,  k  being  a  constant  depending  upon  the  form  of 
the  pillar's  cross-section  (see  Table  III.,  article  62) ; 

/.    Bi  ~  —^ 

Whence  (397)  becomes 


f  and  a  being  constants  to  be  determined  by  experiment  ; 
/.    -  =  —  L  —  ,  (400) 


which  is  of  the  form  "proposed  by  Tredgold,"  and  is  now 
known  as  the  "  Gordon  Formula,"  having  been,  after  some 
"disuse,  revived  by  Mr.  Lewis  Gordon,  who  determined  the 
values  "  of  a  and  /,  for  certain  materials,  from  the  results  of 
Mr.  Hodgkinson's  experiments. 


3OO  MECHANICS  OF  THE   GIRDER. 

1  19.  If,  in  equation  (399),  we  put  Sr2  for  /,  using  still  the 
notation  of  article  115,  we  find 


PDh      P/2 

'~  (40I) 


Therefore,  from  (397), 


(402) 


which  is  Professor  Rankine's  modification  of  the  Gordon  for 
mula  ;  r  being  the  least  radius  of  gyration  of  the  cross-section. 
The  Gordon  (400)  and  the  Rankine  (402)  formulae  are  iden 
tical  if  we  make 


i2o.  Supposing  f  to  be  constant  for  varying  conditions  of 
the  pillar,  both  a  and  al  will  be  found  to  require  different  co- 
efficients, according  as  the  pillar  has  neither,  one,  or  both,  of  its 
ends  fixed. 

Assuming  that  equations  (400)  and  (402)  apply  to  a  pillar 
fixed  in  direction  at  both  ends,  Fig.  109,  we  see  that  the 
length,  /,  between  the  points  of  contrary  flexure,  is  in  the  con- 
dition of  a  pillar  not  fixed  at  its  ends,  and  has  only  the  strength 
of  a  pillar  of  twice  its  length,  2/,  fixed  at  both  ends ;  that  is, 
for  a  pillar  rounded  at  both  ends,  we  have,  — 

Gordon's  formula, 

-r-k  y» 

(404) 


PILLARS.  301 


Rankine's  formula, 

P 

—  =  — 

——  —  —  ^—  . 

(405) 

S 

i 

+  — 

Similarly,  in  Fig.  109,  the  length,  /,  between  either  point  of 
contrary  flexure  and  the  remoter  end  is  in  the  condition  of  a 
pillar  with  one  fixed  and  one  rounded  end,  and  has  only  the 
strength  of  a  pillar  f  /  in  length.  We  have,  then,  for  a  pillar 
fixed  at  one  end  and  rounded  at  the  other,  — 

Gordon's  formula, 

_  L  _  (406) 


S 
Rankine's  formula, 


(407) 


This  is  Mr.  Hodgkinson's  ingenious  explanation  of  the  vari- 
ation among  the  strengths  of  these  three  classes  of  pillars,  a 
variation  which  he  discovered  by  a  comparison  of  the  results  of 
his  experiments. 

If  we  invert  the  three  numerical  co-efficients  of  the  fractions 
in  the  denominators  of  (400),  (404),  (406),  or  (402),  (405),  (407) 
(viz.,  i,  4,  J^-),  and  multiply  the  inverted  numbers  by  4,  we  have 
the  relation,  4,  i,  2.25  ;  while  4,  i,  2.28,  is  the  relation  of  the 
corresponding  constants  in  equations  (385),  (381),  (391),  deter- 
mined analytically.  We  may  hence  infer  that  the  degree  of 
approximation  in  (391)  is  close  to  the  true  value.  Especially, 
since  we  can  seldom  tell  the  exact  amount  of  influence  which 
given  end  bearings  exert,  may  we  regard  (391)  practically  cor- 
rect. 


302 


MECHANICS  OF   THE   GIRDER. 


VALUES  OF  /  AND  a 

A.,  American  Bridge  Company 


TABLE    IV. 

OF  THE  GORDON,  AND  OF/X  AND  al  OF  THE 
RANKINE  FORMULA. 

,  Chicago,  111.;  K.,  Keystone  Bridge  Company,  Pittsburgh,  Penn. 


Material. 

Form 
of 
Section. 

Experi- 
menters. 

Authority. 

Gordon  Formula. 

Rankine  Formula. 

/ 

a 

A 

* 

Iron,  Cast     .     . 

0 

Hodgkinson. 

Gordon. 

80000 

400 

- 

- 

Iron,  Cast     .     . 

O 

Hodgkinson. 

Gordon. 

80000 

267 

- 

- 

Iron,  Cast 

cjja 

Hodgkinson. 

Gordon. 

80000 

133 

- 

- 

Iron,  Wrought  . 

• 

Hodgkinson. 

Gordon. 

36000 

3000 

- 

- 

Iron,  Wrought  . 

ft 

Hodgkinson. 

Rankine. 

36000 

3000 

36000 

36000 

Iron,  Wrought  . 

ei 

Hodgkinson. 

Stoney. 

35840 

3000 

- 

- 

Iron,  Wrought  . 

Hodgkinson. 

Stoney. 

30660 

3000 

- 

- 

Iron,  Wrought  . 

Hodgkinson. 

Stoney. 

40032 

3000 

- 

-  ' 

Iron,  Wrought  . 

LT  E^ 

Davies. 

Unwin. 

42560 

900 

- 

- 

Iron,  Wrought  . 

=  O 

A.        K. 

Lovett. 

49580 

3000 

42980 

36000 

Iron,  Wrought  . 

n 

A.        K. 

Lovett. 

43725 

3000 

38650 

36000 

Iron,  Wrought  . 

3C 

A. 

Lovett. 

38271 

3000 

37029 

36000 

Iron,  Wrought  . 

!3 

K. 

Lovett. 

36523 

3000 

33531 

36000 

Steel,  Mild  .    . 

• 

- 

Baker. 

67200 

1400 

- 

- 

Steel,  Strong     . 

« 

- 

Baker. 

114240 

900 

- 

- 

Steel,  Mild  .     . 

• 

'    - 

Baker. 

67200 

2480 

- 

- 

Steel,  Strong     . 

(C 

- 

Baker. 

114240 

1600 

- 

- 

Timber     .     .     . 

tt 

Hodgkinson. 

Rankine. 

7200 

250 

- 

- 

Oak  and  Fir      . 

« 

Rondelet, 

Stoney. 

1.5  C  of 
Table  II. 

250 

1 

. 

Stone  and  Brick, 

(I 

Rankine. 

Cof 

Table  II. 

600 

- 

- 

PILLARS.  303 


SECTION  4. 

Strength  of  Pillars  computed  by  the  Preceding  Formula,  and  compared 
with  the  Strength  experimentally  determined. 

121.  The  following  tables,  V.,  VL,  VII.,  VIIL,  IX.,  X.,  XII., 

contain  data  derived  from  experiments  on  the  strength  of  pil- 
lars, probably  as  trustworthy  as  any  yet  made  and  published. 
To  these  tests  the  appropriate  formulae,  either  direct  or  in- 
verted, have  been  applied ;  and  the  values  of  /,  C,  or  Q  for  a 
given  pillar,  computed  by  different  formulae,  have  been  tabu- 
lated in  the  same  horizontal  line. 

In  Table  XI.  no  experimental  values  are  given,  but  the 
assumed  values  of  E  and  C  are  within  the  limits  fixed  by 
experiments  upon  steel.  In  Table  VII.,  when  the  thickness,  /, 
of  the  metal  is  less  than  a  fifty-fifth  part  of  the  least  diameter, 
k,  of  the  pillar,  the  computed  value  of  Q,  the  breaking-weight, 
in  pounds,  per  square  inch,  has  been  diminished  in  the  ratio 

-7-,  as  seems  to  be  required  by  the  test^. 

For  columns  having  rounded  or  hinged  ends,  in  Table  V, 
the  formulae  for  those  having  one  flat  and  one  round  end  have 
been  used,  as  more  in  harmony  with  the  tests  than  the  formulae 
for  columns  having  no  end  moments. 

It  must  be  confessed  that  there  are  anomalies  of  consider- 
able magnitude  in  the  experiments  themselves ;  and,  of  course, 
there  appear  corresponding  variations  from  the  test  values  in 
the  numbers  computed  according  to  the  laws  of  the  applied 
formulae. 

It  is  to  be  regretted  that  we  have  not,  accompanying  these 
tests  for  Q,  also  experimental  determinations  of  C  and  of  E,  for 
each  pillar  tabulated,  but  have  been  obliged  to  use  probable 
mean  values  of  C  in  all  the  calculations  of  Q,  and  probable 
mean  values  of  E  in  all  but  Table  V. 


304 


MECHANICS  OF   THE   GIRDER. 


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PILLARS. 


305 


TABLE    VI. 

SOLID  RECTANGULAR  PILLARS  OF  WROUGHT-!RON. 

Flat  ends  well  bedded;    Hodgkinson's  Experiments. 

DATA  FROM   BINDON   B.  STONEY'S   THEORY  OF   STRAINS    IN    GIRDERS   AND 


SIMILAR  STRUCTURES. 

Assume  E  =  27,311,111,  and  C 


50,000. 


S 

i 

/ 

t+M 

*-? 

Excess  over  Q  by 

No. 

Ratio  of 

Breaking- 

Gordon  Formula, 

Equation  (385), 

Sectional 
Area. 

Least 
Diame- 

Length. 

Length  to 
Least 

Weight, 
by 

a-jgf- 

^"x  +   ci* 

ter. 

Diameter. 

Experiment. 

[+  3000*2 

+  4n*£r* 

sq.  ins. 

ins. 

ins. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in. 

30 

1.0465 

1.0230 

7-5 

7-331 

48682 

-13473 

-     134 

31 

1.0465 

1.0230 

15-0 

14-663 

34554 

-   1105 

+  10103 

32 

1-0475 

1.0230 

30.0 

29.326 

25327 

+   2527 

+   8489 

33 

2.9880 

0.9960 

30.0 

30.121 

29655 

-  2137 

+   3570 

34 

2.2970 

0.7630 

30.0 

39-3I9 

27767 

-  4"5 

-     890 

35 

1.0486 

1.0240 

60.0 

58-594 

17268 

-     555 

-       88 

36 

4.5900 

1.5300 

90.0 

58-524 

19987 

-  3343 

_   2896 

37 

1.5011 

0.5026 

30.0 

59-689 

16853 

-     470 

-       90 

38 

5.8166 

0.9960 

60.0 

60.241 

17698 

-   M79 

—  "38 

39 

2.9950 

0.9950 

60.0 

60.303 

18067 

-   1865 

—  1530 

40 

2.3090 

0.7670 

60.0 

78.227 

12969 

—   "79 

—   1619 

4i 

4-5300 

1.5100 

120.0 

79.470 

10165 

+   1377 

+     9'4 

42 

1.0490 

i  .0240 

90.0 

87.891 

9753 

+     273 

-     3i7 

43 

2.9915 

0-9955 

90.0 

90.407 

9912 

-     289 

-     900 

44 

5-8307 

0.9950 

9O.O 

90.452 

9280 

+     336 

-     276 

45 

1.4980 

0.5070 

6o.O 

"8.343 

5653 

+     670 

+       33 

46 

1.5110 

0.5070 

00.0 

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5604 

+     7*9 

+       82 

47 

2.9750 

0.9950 

120.0 

120.603 

4280 

+   1848 

+    1218 

48 

2.3060 

0.7660 

120.0 

156.658 

3379 

+     525 

+       32 

49 

1.4980 

0.5023 

90.0 

179.176 

2410 

-1-     653 

+     240 

50 

1.4980 

0.5030 

120.0 

238.659 

816 

+     978 

+     7H 

The  substance  of  section  i  of  this  chapter,  together  with  some  of  these  tables, 
appeared  first  in  a  contribution  by  the  author  to  Van  Nostrand's  "  Eclectic 
Engineering  Magazine  "  for  December,  1879,  New  York. 

It  is  confidently  expected  that  the  new  United-States  Government  testing- 
machine,  already  in  use  at  the  Arsenal  in  Watertown,  Mass.,  will  contribute  a  set 
of  values  for  the  constants  to  be  used  in  tension,  compression,  cross-breaking,  and 
torsion,  and  for  pillars,  much  more  in  agreement  with  the  capabilities  of  the  actual 
members  of  structures  than  any  values  of  these  constants  (if,  indeed,  they  shall 
turn  out  to  be  constants  at  all)  hitherto  determined. 


306 


MECHANICS  OF   THE   GIRDER. 


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PILLARS. 


307 


TABLE    VIII. 

HOLLOW   CYLINDRICAL  PILLARS   OF  WROUGHT- IRON. 

Ends  flat  and  well  bedded.     Hodgkinson's  Experiments. 

DATA  FROM  BINDON  B.  STONEY'S  "THEORY  OF  STRAINS   IN    GIRDERS   AND. 
SIMILAR  STRUCTURES."      hz  —  8r2. 


No. 

5 

h 

l  +  h 

l  +  r 

h  +  t 

Q  =  Breaking-  Wt.,  in  Lbs.,  per  Sq.  Inch, 

Sec- 
tional 
Area. 

Diame- 
ter. 

Ratio 
of 
Length 
to 
Diame- 
ter. 

Ratio  of 
Length 
to 
Radius 
of 
Gyration. 

Ratio  of 
Diame- 
ter to 
Thick- 
ness of 
Metal. 

*By- 
Experi- 
ment. 

Gordon 
Formula, 

(?=_J00^0_ 

Equation  (385), 

1  +            - 
30oo«2 

I4V*H 

sq.  ins. 

ins. 

80 

0.444 

1-495   , 

80.00 

226.274 

15.00 

M673 

12782 

12873 

81 

0.610 

1.964 

60.00 

172.816 

18.80 

23206 

18204 

18127 

82 

1-435 

2.340 

51.28 

145.042 

10.80 

22179 

21342 

21810 

83 

1.605 

2.350 

51.00 

144.250 

9.70 

21572 

21451 

21926 

84 

0.804 

2.490 

47.80 

I35.I99 

23.27 

29798 

22735 

23289 

85 

0.444 

1-495 

40.00 

"3-137 

15.00 

31180 

26120 

26914' 

86 

i-35° 

3.000 

40.00 

112.877 

20.00 

27671 

26120 

26959 

87 

0.610 

1.964 

30-50 

86.267 

18.80 

33299 

30571 

3^745 

88 

1.414 

3-035 

29.60 

83-721 

18.00 

29789 

30997 

32212 

89 

1.707 

4-050 

29.60 

83.721 

29.00 

27657 

30997 

32212 

90 

1.900 

4.060 

29.60 

83.721 

26.10 

26263 

30997 

32212 

9* 

i-37i 

2-335 

25.70 

72.690 

11.40 

29998 

32823 

34219 

92 

1.472    |    2.350 

25-50 

72.125 

10.60 

29330 

33024 

34321 

93 

0.804 

2.490 

24.10 

68.165 

23-27 

35ioo 

33554 

35025 

94 

1.613 

4.052 

22.20 

62.791 

30.90 

33331 

34399 

3596i 

95 

2.879 

4.000 

22.20 

62.791 

16.50 

26046 

34399 

3596i 

96 

2.897 

4.000 

22.20 

62.791 

16.00 

26503 

34399 

3596i 

97 

2-837 

4.000 

22.2O 

62.791 

16.50 

27816 

34399 

3596i 

98 

0.804 

2.490 

21.  OO 

59-397 

23-27 

36489 

34917 

36536 

99 

0.444 

1-495 

20.00 

56.569 

15.00 

34220 

35338 

37004 

100 

1.800 

6.180 

1940 

54-87I 

65.00 

33375 

35586 

37280 

101 

2.540 

6.360 

18.90 

53-334 

49.00 

35985 

35789 

37524 

1  02 

0.610 

1.964 

I5.30 

43-275 

1  8.80 

36980 

37I5I 

39028 

103 

2.895 

3-995 

15.00 

42.426 

16.30 

30024 

37256 

39*45 

104 

2.848 

3-995 

15.00 

42.426 

16.50 

34453 

37256 

39M5 

105 

2-547 

6.366 

14.10 

39-88i 

48.90 

41664 

3756i 

39487 

106 

1.407 

2-343 

12.80 

36.204 

II.  10 

38214 

37976 

39953 

107 

1-435 

2-335 

12.80 

36.204 

11.40 

36639 

37976 

39953 

108 

1-435 

2-335 

12.80 

36-204 

11.40 

35389 

37976 

39953 

109 

1.651 

2.383 

12.50 

35-355 

9-70 

33107 

38067 

40055 

no 

1-358 

2-343 

12.30 

34-790 

11.60 

39569 

38127  . 

40030 

III 

1-554 

2-373 

12.20 

34-507 

10.27 

36906 

38i57 

40155 

112 

1.799 

6-175 

9-70 

28.075 

61.10 

38355 

38832 

40853 

"3 

1.414 

3.000 

9-30 

26.305 

19.60 

37392 

38928 

41023 

114 

2.845 

4.000 

7.00 

19.799 

16.00 

47844 

39406 

41182 

"5 

2.850 

4.026 

6-95 

19-657 

16.00 

48567 

39415 

4J573 

116 

1.799 

6.125 

4.90 

13-859 

62.50 

41361 

39732 

4i93i 

Assume  E  =  24,000,000. 


C  =  42,290,  f  =  40,050,  Means. 


308 


MECHANICS  OF   THE   GIRDER. 


TABLE    IX. 

SOLID  CYLINDRICAL  PILLARS  OF  CAST-IRON. 

Ends  flat  and  well  bedded.     Hodgkinson's  Experiments. 
DATA,  AND  PER  CENT  OF  VARIATION  FROM  Q,  BY  THE  HODGKINSON  AND  GORDON  FORMUUE, 

TAKEN   FROM   WlLLIAM   E.  MERRILL'S  "  IRON  TRUSS   BRIDGES  FOR  RAILROADS." 

hz  =  i6r2,  E  =  12,215,000,  C  =  109,801,  y=  80,000. 


/+* 

h 

Q 

Variation  from  Q,  per  cent,  by 

Gordon's  For- 

No. 

Ratio 
of 
Length 

Diameter. 

Breaking- 
Weight, 
by 

Hodgkinson's 
Formulae, 
Equations  (392), 

mula, 
Equation  (400), 
80000 

Equation 
(385). 

to 
Diameter. 

Experiment. 

(394). 

ins. 

Ibs.  per  sq.  in. 

117 

4 

0.520 

107674 

-  6 

—  29.000 

-  3-641 

n8 

8 

0.500 

88964 

-  3 

—21.000 

+  0.085 

119 

10 

0.777 

67502  • 

+13 

—   4-000 

+19.227 

120 

13 

0.768 

55959 

+13 

—   O.OOI 

+21.444 

121 

IS 

0.500 

57321 

+    2 

—  II.OOO 

+  5-267 

122 

15 

0.785 

50182 

+U 

+   Won 

+20.243 

I23 

15 

I.  000 

51248 

+  9 

+  7.000 

+I7-74I 

124 

20 

0.500 

45485 

—  i 

—13.000 

-  1.758 

125 

20 

0.775 

45596 

—  i 

•  —  lo.ooo 

-  1.998 

126 

20 

1.022 

38770 

+12 

+  5-000 

+I5-257 

I27 

24 

0.500 

36644 

+    2 

—II.OOO 

-  3-291 

128 

26 

0.777 

32860 

+      \ 

—  9.000 

-  3-503 

I29 

30 

O.5IO 

33"i 

—10 

—  25.000 

-22.497 

I30 

30 

1.  010 

25350 

+  5 

—  3.000 

+  1.231 

131 

39 

0.770 

18921 

-  8 

—13.000 

—11.284 

132 

39 

1.560 

I5I53 

+  6 

+11.000 

+10.777 

133 

40 

0.510 

18749 

—   2 

—  13.000 

-14.241 

134 

47 

1.290 

12291 

-  3 

+  \ 

—  1.261 

135 

61 

0.500 

8464 

+  6 

—  7.000 

—  10.881 

136 

61 

0.997 

7990 

+    \ 

—    2.000 

-  5-581 

137 

79 

0.770 

5274 

+    2 

—  8.000 

-12.286 

138 

119 

0.510 

2384 

+19 

—  7.000 

—  12.416 

Equation  (392),  as  used  in  Table  IX.,  is 
P  =  98922. 


See  "  Iron  Truss  Bridges  for  Railroads,"  p.  26. 

For  E  =  12,215,000,  see  Stoney's  "Theory  of  Strains,"  p.  180. 


PILLARS. 


309 


TABLE    X. 

SOLID  CYLINDRICAL  PILLARS  OF  CAST-IRON. 

Ends  rounded.     Hodgkinson's  Experiments. 
DATA,  AND  PER  CENT  OF  VARIATION  FROM   Q,  BY  THE  HODGKINSON  AND  GORDON   FORMUUE, 

TAKEN  FROM   WlLLIAM   E.  MERRILL'S   "  IRON  TRUSS   BRIDGES  FOR  RAILROADS." 

hz  =  i6r2,  E  =  15,268,750,  C  =  109,801,  /  =  80,000. 


M 

h 

Q 

Variation  from  Q,  per  cent,  by 

No. 

Ratio 
of 

Breaking- 

Hodgkinson's 
Formulae, 

Gordon's  For- 
mula, 

Equation, 

Length 
to 

Diameter. 

Weight, 
by 

Equation  (392), 

2  =  _8oooo_ 

£• 

Diameter. 

Experiment. 

P  —  *m7Q 

T     J_ 

I+^ 

3379dV)'7 

looA* 

ins. 

Ibs.  per  sq.  in. 

i39 

8 

0.500 

76939 

-25 

-34 

—18.269 

140 

10 

0.770 

49280 

-  7 

-i  8 

+    2.877 

141 

13 

0.760 

38590 

-IS 

-25 

—    4-203 

142 

15 

0.497 

27124 

+    £ 

—ii 

+"•735 

i43 

15 

0.990 

25660 

+  10 

-  6 

+18.110 

144 

20 

0.760 

20331 

-13 

—21 

—  4-633 

i45 

20 

1.  010 

19642 

-  9 

_I9 

-  1.288 

146 

20 

1.520 

17928 

+  3 

—10 

+  8.149 

i47 

23 

1.290 

13187 

+  5 

-  7 

+16.175 

148 

26 

0.767 

14289 

-23 

-29 

-I3-472 

149 

3° 

0.500 

9697 

-13 

-19 

—  1.464 

15° 

30 

0.990 

7931 

+  9 

—   2 

+20.477 

151 

31 

1.940 

7717 

+13 

-  3 

+16.599 

152 

31 

1.960 

8051 

+14 

-  3 

+"•763 

i53 

34 

1-765 

6360 

+  5 

—10 

+19.262 

i54 

34 

1.780 

7058 

+  6 

—  IO 

+  7-467 

i55 

39 

0.770 

5854 

-  5 

-17 

+  0.137 

156 

39 

1-535 

5755 

+    * 

-16 

+  1-859 

i57 

40 

1.520 

5985 

-    i 

—14 

—  6.650 

158 

47 

1.290 

4367 

_    £ 

-18 

—  6.000 

i59 

47 

1-295 

4149 

~~      3 

-18 

—  i.  060 

160 

61 

0.500 

2745 

-  5 

-23 

-  9.872 

161 

61 

0.990 

2471 

+  8 

-16 

+    O.I2I 

162 

79 

0.770 

1675 

+    2 

-24 

—II.I05 

163 

121 

0.500 

728 

+  10 

-25 

-12.083 

3io 


MECHANICS  OF  THE   GIRDER. 


E 
C 

Eq 


NIOOO- 


O     M     rotxosvo 
r-NwO>-4U">O 


rOHNVOtxCO 

1000     ^-miH     t^ 
Wior^ot-tvo 

000 


Ooo     rooo     O 
ow     oiotx 


8s      ?      M 


OJ 


•s 

«   S 


s  llsf  g 

'  *rl 


a  8 


33333 

ooooo 
K  ff  ff  H  M 


333333333 
bJOMMMbCbOMMbJ) 
CCGCCCCCC 

22SS52252 


OO 
VO      t^ 


PILLARS. 


TABLE    XII. 

.    SOLID  SQUARE  PILLARS  OF  PINE. 

DATA  FROM   BINDON   B.  STONEY'S  "THEORY  OF  STRAINS  IN   GIRDERS  AND 
SIMILAR  STRUCTURES." 

kz  =  i2T2.  Take  E  =  1460000,  C  =  f  =  5000. 


l  +  k 

Q  =  Breaking-  Weight,  in  Lbs.,  per  Square  Inch. 

No. 

Ratio  of 
Length 
to 

Rondelet's 

Brereton's 
Tests. 

Gordon 
Formula, 

Hodgkinson's 
Formula, 

Equation 

Equation 
(391). 

Equation 
(385). 

Least 
Diam- 
eter. 

Propor- 
tionals. 
Flat  Ends. 

Ends  in 
Ordinary 
Manner. 

I+2loA2 

SooC/2 

No  End 
Moment. 

One  End 
fully 
fixed. 

Both 
Ends 
fully 
fixed. 

hz 

184 

i 

5000 

_ 

_ 

5000 

_ 

_ 

_ 

185 

12 

4167 

- 

3176 

5000 

3126 

3959 

4349 

1  86 

24 

2500 

- 

1513 

4940 

1471 

2437 

3126 

187 

36 

1667 

- 

809 

1929 

782 

1485 

2135 

188 

48 

833 

- 

489 

1085 

462 

960 

1471 

189 

60 

4J7 

- 

325 

693 

313 

660 

1076 

190 

72 

209 

- 

230 

483 

221 

478 

642 

191 

10 

- 

1867 

3571 

5000 

3530 

4228 

4529 

192 

20 

- 

1789 

1923 

5000 

l876 

2889 

3530 

193 

30 

- 

1400 

1087 

2777 

1053 

1891 

2581 

194 

40 

I 

1244 

676 

1563 

653 

1273 

1875 

312  MECHANICS  OF  THE   GIRDER. 


CHAPTER  VIII. 

PROPORTIONS   AND   WEIGHTS    OF  ALL  THE   MEMBERS    OF  A  BRIDGE 
EXCEPTING   THE   GIRDERS   PROPER. 

122.  The  Floor. 

Let  /  =  length  of  floor,  in  feet. 
q  =  breadth  of  floor,  in  feet. 
t  =  thickness  of  floor,  in  feet. 
u  —  weight  of  one  cubic  foot  of  the  material,  in  pounds. 

.'.    Volume  of  floor  =  Iqt  cubic  feet 

=  o.o  1 2lqt  thousand  feet,  board  measure. 

F  —  weight  of  floor  =  ulqt  pounds.  (408) 

123.  The  Joists,  Longitudinal. 

/  -7-  n  =  length  of  joist  in  each  panel,  in  feet. 
d  =  depth  of  joist,  in  inches. 
b  —  thickness  of  joist,  in  inches. 
n  —  number  of  equal  panels. 
g  =  distance  between  centres  of  joists,  in  feet. 
q  -r-  g  =  number  of  joists  in  any  panel,  each  of  the  two  out- 
side ones  having   the   thickness   \b,   and   being 
counted  as  one-half  a  joist. 

ncl  ~^~  g  —  whole  number  of  joists  in  the  n  panels. 
L  =  panel  weight  of  uniform  load,  in  tons. 
«x  =  weight  of  one  cubic  foot  of  the  material,  in  pounds. 


PROPORTIOATSy   ETC.,    OF  MEMBERS  OF  A   BRIDGE.       313 


Weight  upon  the  joists  of  one  panel       = h  20ooZ  pounds, 

n 


*  tf 

ultg       2  ooogL 
Uniformly  distributed  load  on  one  joist  = 1 pounds. 

Add  weight  of  joist  itself  =  -    —  pounds. 

Total  uniform  load  for  each  joist  is,  therefore, 
ultg       2OoogL       bdfai       I 


where  w  is  the  number  of  pounds  per  linear  foot  to  be  sup- 
ported by  one  joist. 

Now,  by  equation  (52),  we  have  for  the  external  forces, 
greatest  moment  at  centre, 


//\2         Iw       I       ul2tg  i 

M  =  =X=-+--  +  foot-pounds  ; 


and   for  the  internal   forces  of  a  rectangular  beam,  equation 
(160),  the  moment  of  resistance  is 

R  =  \Bbdz  inch-pounds  =  -faBbd2  foot-pounds. 

Introducing  /,  the  factor  of  safety,  and  equating  M  and 
R  -7-  f,  we  find 


ulztg 

(409) 


Taking  the  value  of  B  from  Table  II.,  and  assigning  a  value 
to  b  or  d,  we  may  find,  from  (409),  the  required  depth  or  thick- 
ness of  each  joist. 


3 14  MECHANICS  OF   THE   GIRDER. 

If  we  neglect  the  weight  of  the  joist  itself,  which  omission 
the  factor  of  safety  may  well  warrant,  the  last  term  in  (409) 
vanishes,  and  we  have  at  once 

Thickness  of  joist  =  b  —          2    (uqlt  -f  2Ooo«Z). 

Depth  of  joist         =  d  =  \  -^-(uqlt  -f  2OOO*£)l*. 

( tfqbB  j 


/  =  weight  of  (nq  -*-  g)  joists  =    i      '  pounds.     (410) 

In  a  similar  manner  may  the  dimensions  and  weight  of  any 
other  joist  or  beam  or  stringer  be  found ;  that  is,  by  equating 
the  greatest  moment  due  the  external  forces  acting  on  the 
beam,  to  the  greatest  allowable  moment  due  the  internal  forces 
resisting. 

124.    The    Wrought-Iron    I   Floor    Beams,  Transverse, 
supporting  the  Joists,  Floor,  and  Load. 
Let  dz  =  depth  of  beam,  in  inches. 

dj.  —  depth  of  web,  in  inches. 
d2  —  di  =  depth  of  two  flanges,  in  inches. 

b2  =  breadth  of 'one  flange. 
b2  —  bi  =  thickness  of  web. 

qT  =  entire  length  of  beam,  in  feet. 
5  =  cross-section  of  beam,  in  square  inches. 
n  —  i  =  number  of  beams  in  bridge. 

m  =  weight    of    one   cubic   inch   of    wrought-iron,    in 
pounds. 

F  +  J  +  20OonL 
D  = =  uniform  load  on  any  beam, 

in  pounds. 
Then,  by  equation  (52), 

Moment  of  external  forces  =  M  =  \Dq^  inch-pounds. 


PROPORTIONS,   ETC.,   OF  MEMBERS  OF  A   BRIDGE.       315 

And,  from  equation  (161), 

Moment  of  internal  forces  =  R  =  B^b*d*    ~  b*d^  inch-pounds. 

6d2 

Whence,  introducing  /  as  the  factor  of  safety, 


Let  us  take  now  the  dimensions  of  the  cross-section  of  a 
well-proportioned  I-beam,  as,  for  instance,  d2  —  15,  d*  =  I2j, 
bi  —  5|,  bi  =  4|,  and  express  the  relation 


Therefore  (411)  becomes 


/.  4  =  3.80122 

Area  of  section  =  S  =  bA  —  £,</,  =  rz$j42 

(413) 


=  weight  of  floor  beams  =  \2qjn(n  —  i)S 


3l6  MECHANICS  OF  THE   GIRDER. 

If  the  beam  actually  used  has  a  form  of  cross-section  varying 
materially  from  that  here  assumed,  the  co-efficient  of  (412)  must 
be  made  to  conform  thereto. 

We  may  compensate  for  the  omission  of  the  beam's  own 
weight  from  the  formula,  first,  by  selecting  from  the  manufac- 
turer's list  of  beams  that  one  whose  depth  agrees  most  nearly 
with  our  computed  depth  above  it ;  and  second,  by  using,  in 
the  calculation,  the  entire  length  of  beam,  instead  of  the  net 
length  between  bearings. 

Having  thus  employed  the  formula  to  determine  the  depth 
of  beam  required  for  the  given  load,  the  weight  may  be  taken 
from  the  manufacturer's  tables.  Indeed,  the  manufacturer's 
tables  of  strength  may  be  used  without  this  calculation,  when- 
ever they  are  known  to  be  trustworthy,  by  selecting  the  depth  of 
beam  corresponding  to  the  required  length  and  "  safe  load." 

125.  The  System  of  Lateral  Support.  —  This  system 
includes  whatever  arrangement  of  struts,  ties,  and  braces  is 
employed  to  prevent  a  lateral  bending  of  the  girders,  and  their 
rotation  about  their  points  of  support. 

The  arrangement  must  manifestly  vary  with  the  form  and 
height  of  girder  ;  a  high  girder  with  straight  chords  allowing  a 
complete  horizontal  trussing  overhead  and  under  the  floor,  while 
arched  top  chords  allow  only  a  partial  head-bracing,  and  low 
girders  for  "  through "  bridges  can  only  be  laterally  braced 
from  below. 

In  all  cases,  the  horizontal  systems,  top  and  bottom,  should 
be  rigidly  connected  with  the  girders,  whether  angle  braces  are 
employed  or  not.  For  high  girders  with  straight  chords,  there 
are  generally  used,  a  strut  at  every  pair  of  opposite  top  joints, 
n  +  i  in  number,  and  a  pair  of  diagonal  ties  at  the  top  and  bot- 
tom of  each  panel,  472  in  number. 

The  proportions  of  these  members  may  be  computed  in  the 
same  manner  as  the  proportions  are  found  for  a  girder  uni- 
formly loaded,  using  the  assumed  pressure  of  wind  against  the 


PROPORTIONS,   ETC.,    OF  MEMBERS  OF  A   BRIDGE.       317 

side  of  the  bridge  and   load   as  the  uniform  horizontally  (or 
otherwise)  acting  load. 

For  girders  admitting  full  head-bracing,  we  thus  compute : 

q  =  length  of  horizontal  strut,  in  feet. 
\  <?  +  (-  )  =  length  of  horizontal  diagonal,  in  feet. 

6*r  =  cross-section  of  each  strut,  in  square  inches,  j    Assumed 
S2  =  cross-section  of  each  diagonal,  in  square  >•         or 

inches,  J  computed. 

m  =  weight  of  one  cubic  inch  of  wrought- iron. 

U  =  weight  of  horizontal  struts         =  \2qrn (n  +  i)St.         (415) 
X  =  weight  of  horizontal  diagonals  =  ^SmnS2i/ q2  -\ — -.      (416) 

126.  Finally,  there  should  be  added  whatever  weight  of  wood 
or  iron  is  not  included  in  the  foregoing  specifications,  but  is 
employed  in  the  actual  completion  and  equipment  of  the  struc- 
ture.    Call  this  weight  p  pounds  to  the  panel ;  then  we  have 

Y  =  weight  of  residue  =  np  pounds.  (41?) 

127.  Take  K  =  weight  of  bridge  exclusive  of  the  girders, 
in  pounds ;  then 

K  =  F+J  +  P  +  U  +  X  +  Y  pounds.     (418) 
And  if  G  —  weight  of  girders,  in  pounds, 

Weight  of  bridge  =  2OQQnW  =  K  +  G  pounds.  (419) 


MECHANICS  OF  THE   GIRDER. 


CHAPTER    IX. 

OPEN   GIRDERS   WITH   EQUAL   AND    PARALLEL   STRAIGHT   CHORDS. 

CLASS    IX. 

SECTION  i. 

The  Pratt  Truss  of  Single  System  and  Uniform  Live  Load.  —  Wind 

Pressure. 

128.  Strains  in  Terms  of  the  Structure's  Unknown 
Weight.  —  Let  Fig.  1 1 1  represent  a  girder,  or  built  beam, 
having  a  discontinuous  or  open  web,  and  its  flanges  or  chords 

A  abcdefahB 


<H 


FIG.  in. 


in  straight  horizontal  lines,  AB,  CD.  Let  the  vertical  mem- 
bers, or  posts,  be  in  compression,  and  the  inclined  members,  or 
diagonals,  be  in  tension.  We  then  have  a  girder  which  has 
been  called  the  Murphy,  Whipple,  or  Pratt  truss  of  single 
intersection,  since  the  diagonals  traverse  but  a  single  panel  or 
division  of  the  girder. 


GIRDERS   WITH  EQUAL  AND  PARALLEL   CHORDS.      319 

Take  n  =  number  of  panels. 

/  =  length  of  girder,  in  feet. 
/  -f-  n  =  c  =  length  of  one  panel. 

h  =  height  of  girder  between  centres  of  chords,  in 

feet. 
<f>  =  inclination  of  diagonals  to  chords. 


nh 

I   '  \lr2     -4-     /72  \//-2     -f    k2 


=  length  of  a  diagonal. 


sin  </>       cos  <£       «  cos 
'c2  -f  ^2  =  1  /—  +  h2  =  -V/2  •+-  n2h2  —  length  of  a  diagonal. 


Assume  the  entire  weight  of  the  structure  supported  by  the 
girder,  including  the  girder's  own  weight,  to  be  uniform  through- 
out, and  equal  to  n  W  tons  applied  at  the  lower  joints  ;  viz.,  \  W 
at  C,  \  W  at  D,  and  W  at  each  of  the  other  n  —  i  joints, 
apices,  or  panel  points,  a,,  b»  c»  etc.  W  is  called  the  panel 
weight,  or  apex  load,  due  to  the  permanent  weight  of  the  struc- 
ture. 

Total  pressure  at  C  or  D  =  \n  W  =  resistance  of  pier  to 
the  permanent  load. 

Assume  also  a  uniform  moving-load,  nL,  advancing  by  apex 
loads,  Z,  from  left  to  right,  upon  and  over  the  girder.  We 
then  have  total  weight  =  n(  W  +  L)  tons  ;  weight  at  each 
apex  =  W  +  L  tons  when  fully  loaded. 

With  these  data,  we  proceed  to  find  the  greatest  strains 
developed  in  each  member  of  the  girder  by  the  permanent 
load  n  W,  and  the  uniform  moving-load  nL. 


32O  MECHANICS   OF  THE   GIRDER. 

_  I  _ 

(a)   To  find  the  moment  at  each  joint  due  the  entire  weight 
n(W  -f-  L)>  and  thence  the  horizontal  strain  in  chords  by  equation 

(95). 

H  —  M  -f-  h  =  moment  divided  by  height. 

Equation  (65)  applies  here  if  for  W  we  put  W  -f-  L,  and  we 
have 


(W  , 

-  ^—  -(n  -  i)  X  i, 

(W  -f  L)l, 

.'.    Ha  =  -      -  (n  —  i)  x   i  =  strain  on  Aa,  a  A  ; 


, 

Mb  =  —  tz~(*  -  2)  x  2> 

.'.     Z?3  =  -  ^^  -  («  —  2)  X  2  =  strain  on  ab,  blcl  ; 

(W  +L)l 

^c  =  -  -  ^-^(n  -  3)  X  3, 

'*•     Hc  =  -  ~      -  {n  ~  3)  X  3  =  strain  on  be,  c,d,  ; 


Md  =  -    -  ^n  ~  4)  X  4, 


(W  +  Z)/ 

(n  —  4)  X  4  =  strain  on  cd,  d1e1  \ 


Mh=(-V-^\n  -(«-!)](«  -i), 

(W  +  Z)/ 
.*.    Zfo  =  ^^ [«  —  (#  —  i)]  («  —  i)  =  strain  on  hB,  g^h^ ; 

where  H  is  the  greatest  horizontal  strain,  in  tons,  at  the  suc- 
cessive joints ;  the  strain  on  each  chord  being  assumed  to  act 
at  the  centre  or  axis  of  the  chord,  whose  depth  is  small  com- 
pared with  h. 


GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      $21 

(b)  To  find  the  compression  on  verticals,  and  the  tension  on 
diagonals,  due  to  permanent  load,  nW,  alone. 

From  equation  (65),  dividing  by  h,  and  from  the  formulae 
for  Class  IX.,  article  49, 

HA  =  o; 

W7 
Ha  =   ™L(n  -  i)  X   i, 

2/2  n 

Wl 
=  Ha  —  HA  —  —  :  (n  —  i)  =  hor.  component  of  Aa^^  ; 


—  Ha  =  -  —  (n  —  3)  =  hor.  component  of  abl  ;, 


.'.    Hc  —  Hb  =  -—  (n  —  5)  =  hor.  component  of  bcl  ; 
2.nn 


TT      rrt'   r        /        \  -i  /  \ 

•*•    HB  —  Hh  —  — -     —  (n  —  i)  I  =  hor.  component 

Therefore,  from  the  triangle  of  forces,  equations   (3),  the 
vertical  components  are 

nh 
I 

.'.    ZA  =  \W(n  —  i)  =  compression  on  AC  or  BD, 
Za  =  \W(n  —  3)  =  compression  on  aat  or  hhlt 
Zb   —  \W(n  —  5)  =  compression  on  bb^  or  gglt 

ZB  =  \W(n  —  i)  =  compression  on  BD  or 


322  MECHANICS   OF   THE   GIRDER. 

And  the  strain  Y  along  any  diagonal  is 

&H  -4-  COS  <£, 

or 


~  -  ^  *  «.         +     722^2 

Z  -i-  sin  6  =  — - — — 

nh 


= (n  —  i)  =  tension  on  Aal  or  .%, 


W 

Ya    =  — : (n  —  3)  =  tension  on  abl  or  hgl9 

2  sin  d> 


=  — : — -(n  —  5)  =  tension  on  bc^  or  gft, 

2  sm 


—  (w  —  i)     =  tension  on  ^x  or 


(c)  Maximum  strain  on  verticals  and  diagonals  from  moving- 
load,  nL,  alone. 

To  find  this  strain  ZL,  we  subtract  equation  (64)  from  (68), 
divide  remainder  by  h  for  greatest  difference  of  horizontal 
strains  at  adjacent  joints,  and  multiply  the  quotient  by  tan  <£  ; 
thus,  after  putting  Z  for  W,  the  difference  between  (68)  and 
(64)  is 

—  X  r(r  H-  i)  =  maximum  &Hh  (say), 


x  2.  x  r(r  +  i)  =  ^  X  lill^L),      (422) 
h          2n2  n  2 

where  r  is  the  number  of  apex  loads  on  the  girder  as  the  mov- 
ing-load advances,  and  ZL  is  the  compression  on  the  (r  +  i)th 
vertical  ; 


GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      323 


Z&    =  -   X  i  =  compression  on  bblt  r  —  i  ; 

n 

T 

Zc    —  —  X  3  =  compression  on  ccly  r  —  2  ; 

n 

Zd  =  —  X  6  =  compression  on  */</„  r  =  3  ; 

n 

Ze    =  —  x  10  =  compression  on  eett  r  =  4  ; 


ZB  =  —  *  -  =  compression  on  BD,     r  =  n  —  i. 


The  greatest  strain  on  diagonals  due  to  moving-load,  nL> 
is 

y  =  ZL  -v-  sin<£;  (423) 

Yb  =  -  X  i  =  tension  on  aj,  r  =  i  ; 


Yc   =  -   X  3  =  tension  on  bj,  r  =  2  ; 


X  6  =  tension  on  c^d,  r  =  3  ; 


x  10  =  tension  on  dte,  r  —  4  ; 


L  (n  —  i)n 

YB  =       .     ,  x  -  =  tension  on  h^B,    r  =  n  —  i. 

(d)  Combining  the  strains  due  n  W  and  nL,  and,  for  conven- 
ience, writing  N  for  -^Z)/,  we  find,  for  any  number  of 
panels  :  — 


324 


MECHANICS  OF   THE   GIRDER. 


MAXIMA  STRAINS  IN  PRATT  TRUSS. 


(a— 1)  N. 


2  (n-2)  N. 


3  (n-8)  N. 


i  (n— 4)  N. 


(u-1)  N.  2  (n-2)  N. 

FIG.  112. 


3  (n-3)  N. 


UNIFORM  DEAD  AND  LIVE  LOADS. 

Loads  applied  at  lower  joints  :  — 
W  =.  panel  weight  of  dead  load. 

L  =  panel  weight  of  live  load. 

/  —  length  of  truss  from  centre  to  centre  of  end  pins. 

h  —  height  of  truss  from  centre  to  centre  of  pins. 

n  =  number  of  panels. 


nh 


V/2  -f  n2h2 


N  = 


W 


2nh 


2  tan  <f> 


129.  Weight  of  the  Structure  determined.  —  (a)  To  find 
the  weight  of  the  top  chord. 

Suppose  <2  -f-  /  to  be  the  greatest  allowable  pressure  to  the 
square  inch  of  section  of  top  chord,  and  Q  to  be  of  the  same 
denomination  with  W  and  L  ;  and  suppose  /to  be  a  number 
called  the  factor  of  safety.  Q  is  known  as  the  breaking-weight 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      325 

of  a  column  of  the  given  material,  having  the  length  of  one 
panel,  and  the  cross-section  of  the  top  chord  for  any  given 
panel. 

Let  m  =  weight  of  one  cubic  inch  of  the  material,  in 
pounds. 

We  then  have,  for  each  one  of  the  equal  panel  lengths  of 
the  top  chord, 

Area  of  section  =  J~jL  square  inches, 

Volume  of  one  panel  length  =  ^  —  cubic  inches, 

Qn 

Weight  of  one  panel  length    =  I2mfl  H  pounds, 

Qn 


Weight  of  top  chord  =  ^H  pounds. 

Qn 

From  (a}  of  the  preceding  article  we  find 

«[i  +  2  -I-  3  -f  4  -f  .  .  .  (n  —  i)  terms] 
-  [i2  +  2*  +  3*  +  4*  +  .  .  .  (rc  -  i)  terms] 
-f-  i#2  for  n  even,     +  J(#2  ~~  i)  for  w  odd.    j 


T\7 


znh 

(W  +  L)l 

*  -  ^     ' 


X  — (2n2  +  3»  —  2),  n  even; 

12 


X    A(2«3    +   3^   _    2«   _   3),  „  Odd. 


Substituting  these  values  for  ^H,  we  have 

Weight  of  top  chord  =  mfl*(w  +  L\2n*  +  -„  _  2) 

2Qnh 

(n  even), 

(424) 


(wodd). 


326  MECHANICS  OF  THE   GIRDER. 

(b)   Similarly,  if   T  -f-  /  —  the  greatest  allowable  tensile 
strain,  we  find 


Weight  of  bottom  chord  =  -^H  pounds. 


4-  2(n  —  i)  for  two  end  panels, 
n\_i  +  2  +  3  +  4  +  .  .  .  (n  —  i)  terms] 

-  [i2  +  22  4-  32  +  ..-(»-  i)  terms] 
—  J«2  for  «  even,     —  %(n2  —  i)  for  n  odd. 


znh 


=  ± — — i-  x  -rV(2w3  "~  3^2  +  22«  —  24)  when  n  is  even, 

—  21)  when  n  is  odd. 


/.     Weight  of  bottom  chord 


0  even), 


-  3^          22«    -   21) 

(«odd). 


(425) 


(c)  In  finding  the  weight  of  the  verticals,  let  Qt  -f-  /  be  the 
allowed  working  unit  strain  in  compression  ;  then 

.  Area  of  section  =         W  Q —  square  inches, 


Volume  of  one  strut     =  -   ^—       -  cubic  inches, 

\2mfh(Zw  4-  ZL) 
Weight  of  one  strut      = -Q pounds, 

Weight  of  all  verticals  =  -      -Q pounds. 


GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      327 


Hence,  from  the  strain  sheet,  Fig.  112,  using  the  proper  limits 
of  summation,  we  derive,  when  n  is  even, 


=  2  x  — 

2 

_  Wn2 
—          ) 
4 


---  (J»)  terms]  \ 


=  2  x  — 


!(J»)  -  »[i  +  3  +  5  +  7  .  .  .  (%n)  terms] 
[~i  +  3  +  6  +  10  +  .  .  .  f-  -  i\  terms! 

+  if  n  -  -}[n  -  (  -  +  i  Y]   for  middle  strut  I 


S>-)} 


But  when  «  is  odd,  we  thus  sum, 

=  2X«:j^    /I  +  3 


2  x  — N2 


2  \  2 

+  2(  i  +  3  +  6  +  10  +  .  .  .  '^-5  terms 

£ 

=  —  X  ^V\7^3  ~~  3^2  ~  7W  +  3). 
^ 

Wherefore 

3;«/%  ^F«2       w/7?Z 
Weight  of  verticals  =  ~^7\ H  -J<T~(ln2  +  3n  ~  IO) 

(«  even), 


3) 


(«odd). 


(426) 


328  MECHANICS  OF   THE   GIRDER. 

(d)  In  determining  the  weight  of  the  diagonals  in  terms 
of  the  unknown  weight  of  the  structure,  n  W,  we  shall  disregard 
the  effect  of  the  permanent  weight,  n  W,  upon  the  strains 
developed  in  the  counter  diagonals  by  the  moving-load,  nL. 

By  so  doing,  the  value  of  W  comes  out  a  little  greater  than 
strict  theory  requires  ;  but  in  general  practice  the  "  counters  " 
are  inserted  somewhat  in  excess  of  theoretical  demands. 

When,  however,  W  shall  have  been  thus  determined,  the 
strains  upon  all  the  members  are  to  be  computed  according  to 
the  strain  sheet,  Fig.  1 1 2. 

Strain  on  any  diagonal  due  to  L  is  YL. 

Area  of  cross-section        =  £-~t  square  inches, 
Volume  of  one  diagonal  =     T  2/ l  YL  cubic  inches, 

Weight  of  one  diagonal    =  if  .      YL  pounds. 

T  sin  (ft 

From  Fig.  112, 

=  2  x  — : — -[i  +  3  +  6  +  10 +  ...(«  —  i )  terms] 


• 
n  sin  (f>  6 

therefore  weight  of   diagonals  due  uniform  moving-load,  nL, 
alone  is 

*  -  '>•          <**> 


The  weight  of  the  diagonals  due  to  the  dead  load,  n  W,  is 
manifestly  to  be  derived  from  the  weight  of  the  verticals  due 
dead  load  if  for  h  we  put  h  -f-  sin2  0,  and  for  Ql  we  put  T. 


GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      329 


.-.     Weight  of  diagonals  =  ^fhL("  -  i)  + 

2  Tsm2<f> 


(n  even), 

= 


j> 
(*odd). 

(e)  Taking  the  sum  of  the  weights  thus  found,  we  have,  when 
n  is  even,  total  weight  of  girder,  in  pounds, 

„    _   mflz(W  4-   Z)/2/Z2    +    3^    —    2      , 


Q  Tn 


Atell        (429) 


But  when  n  is  0dk/,  total  weight  of  girder,  in  pounds,  is 

2 

Q 


3?*2—  zn  —  3       2^  —  3^  +  22^  —  2i 


+  ,nfhL\  V»  ~  &  '-  V*  +  3  +  A*gL=j)  ).      (43o) 
(  2  ^«  2^sm2</>    ) 

EXAMPLE  i.  —  Wrought-iron  girder  of  6  equal  panels.  Take 
w  =  6,  /  =  60  feet,  //  =  10  feet,  /  =  4,  ?/z  =  y5^  pound,  L  =  8 
tons,  7"  =  24  tons,  g  =  16  tons,  (2i  =  I2  tons,  tan  </>  =  I, 
sin  <^>  =  ^2  =  0.70711.  Therefore,  from  (429), 

£  =  483-333^  +  4267  pounds, 
equal  to  2000/2  W  if  n  W  is  the  girder's  own  weight  in  tons. 

/.     Panel  weight  of  girder  =   W    —  0.3704775  ton, 
Total  weight  of  girder  =  nW  =  2.2228650  tons. 


330  MECHANICS  OF  THE   GIRDER. 

130.  But  if  the  structure  is  a  bridge  having  two  equal  gir- 
ders whose  combined  weight  is  G,  and  an  additional  permanent 
weight  of  K  pounds,  then  the  weight  of  the  bridge  is 

20ocm  W  =  K  +  G  pounds, 

as  shown  by  equation  (419). 

Continuing  the  first  example  of  article  129,  we  compute  K 
as  follows  :  — 

For  the  floor,  we  have  /  =  60  feet  =  length. 

Take  q  —  18  feet  =  breadth. 

t  =  —  feet  =  thickness. 

u  =  54  pounds  =  weight  of  one  cubic  foot  of  oak. 

From  (408), 
Weight  of  floor  =  54  X  60  x  18  X  -jj  =  12150  pounds  =  F. 

The  joists  :  — 

/  -5-  n  =  10  feet  =  panel  length  of  joist. 

Take  b  =  3  inches  =  thickness. 

g  =  2  feet  —  space  between  centres. 
q  4-  g  =.  9  =  number  of  joists  in  each  panel. 
nq  -T-  g  =  54  =  number  of  joists  in  bridge. 
«,  =  54  =  «. 
B  =  10,600. 

/  =  <* 

Then,  by  article  123,  we  have 

d=\      9X9X60x2     /      x  I8  x  6o  x  M+  200o  X  6  X 
(62  X  18  X  3  X  io6oo\  12 

=  7.1424  inches. 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      331 

Call  d  =  8  inches, 

3  X  8  X  60  X   18  X  54 
.-.    Weight  of  joists  = -— =  4860  pounds  =  J. 

For  the  iron  I-beams,  we  have,  from  article  124, 

F  4-  /  H-  2QoonL 
D  — =  18835  pounds. 

Take  qt  =  19  feet  =  entire  length  of  beam. 
f  —  4  =  factor  of  safety. 
B  =  52,567,  from  Table  II. 

Whence,  by  equation  (412), 

Required  depth  of  beam  =  d2  =  3.8oi22f 7 ) 

=  11.435  inches. 
Call  d2  =  12  inches ;  then,  by  (413), 

Area  of  section  =  S  =  r.»M3r/l8835  X  19  X  4V  x  /_IL_Y 

\          52567         /        \"-435/ 
=  12.84  square  inches, 

since  similar  sections  are  to  each  other  as  the  squares  of  their 
like  dimensions. 

Now  this  cross-section,  12.84,  agrees  very  nearly  with  that 
of  the  "12-inch  light  I-beam"  of  the  Union  Iron  Mills,  Pitts- 
burgh, Penn.,  whose  weight  is  42  pounds  to  the  foot,  and 
area  =  42  X  -^  —  12.6  square  inches. 

Using  this  beam,  we  then  have 

Weight  of  5  floor  beams  =  5x19X42  =  3990  pounds  =  P. 

Use  full  head  trussing ;  the  struts  to  be  composed  of  two 
T-bars,  each  5}  pounds  to  the  foot,  latticed  with  i|  X  J  inch 


332  MECHANICS  OF   THE   GIRDER. 

braces,  at  45  degrees,  the  whole  weighing  12 \  pounds  to  the 
running  foot ;  length  =  18  feet. 

Weight  of  (n  +  i)  horizontal  struts  =  7X18X12^ 

=  1575  pounds  =  U. 

Let  the  horizontal  diagonal  ties  be  ij  inches  in  diameter, 
weighing  3.359  pounds  to  the  foot.     Then 


Weight  of  24  horizontal  ties  =  24  x  3-35  9V  io2  +  i82 

=  1660  pounds  =  X. 

Call   the   residue    100  pounds   to   the    panel  ;    that   is,    in 
all  =  600  pounds  —  Y. 

.\     K=F  +  J  +  P  +  U  +  X  +  Y=  24835  pounds, 

G  =  weight  of  girders  =  4267  +  483.333^, 
K  +  G  =  weight  of  bridge  =  29102  +  483.333^ 

=  1  2000  W  pounds  ; 
/.     Panel  weight  of  bridge  =  W  =  2.526947  tons, 

Total  weight  of  bridge  =  nW  =  15.161682  tons. 
Panel  weight  of  dead  load  on  each  girder  =  1.26347  tons, 
Panel  weight  of  live  load  on  each  girder    =  4  tons. 

+  L)  =  5.26347  tons  =  total  panel  weight  for  one  girder. 


Putting  this  value,  5.26347  tons,  for  W  +  L,  in  the  expres- 
sion for  N,  article  128,  (d\  we  find 


c.  26347  X  60 

N  =  -  —  -T-  =       J*  J  v  -  =  2.63174  tons. 
2nh  2  X  6  X   io 

And  from  the  strain  sheet,  Fig.  112,  the  greatest  chord  strains 

are 

fft  =  2.63174  x  5  X  i  =  13.15870  tons, 

H2  =  2.63174  x  4  X  2  =  21.05392  tons, 

Hz  =  2.63174  X  3  X  3  =  23.68566  tons. 


GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      333 


Putting  |  W  =  1.26347  for  W,  and  \L  —  4  for  L,  the  same 
strain  sheet  gives,  for  each  of  two  girders  :  — 
Greatest  compression  on  verticals  : 

Z,  —  0.63174  x  5  -h  0.33333  x  5  X  6  =  13.15870  tons, 

Z2  =  0.63174  x  3  H-  0.33333  x  4  X  5  =     8.56188  tons, 

Z3  =  0.63174  x  i  +  0.33333  x  3  X  4  =    4-63174  tons, 

Z4  =  0.33333  x  2  x  3  =     2.00000  tons. 


Also,  for  the  diagonals  : 
L 


becomes 


=  0.94281  ton, 


2  sin  <£ 


becomes 


0-63T74  =  0.89342  ton. 


sin 


And,  from  Fig.  112  :  — 

Greatest  strain  on  diagonals  : 


Ylt  counter,  =  o 

Y2)  counter,  =  0.94281  X 

Y3,  counter,  =  0.94281  x 

Y4,  main,       =  0.94281  x 

Ys,  main,       =  0.94281  X 

Y6)  main,       =  0.94281  x 

GREATEST  STRAINS,  IN  TONS. 


-  0.89342  x  5  < 

i  —  0.89342  x  3  < 

3  -  0.89342  x  i  = 

6  H-  0.89342  x  i  = 

10  -f  0.89342  x  3  =  12.10836  tons; 

15  -f-  0.89342  x  5  =  18.60925  tons. 


o     ton; 
o     ton; 
1.93501  tons; 
6.55028  tons; 


REQUIRED  SECTIONS,  IN  SQUARE  INCHES. 


LOAD  APPLIED  AT  BOTTOM  JOINTS.     DIAGONALS  IN  TENSION.  —  CLASS  IX. 
FIG.  113. 

131.  Now,  it  is  very  evident  that  the  bridge  would  depend 
entirely  upon  the  floor  system  for  stability  in  a  longitudinal 
direction  if  we  should  omit  the  bottom  chords  and  the  counter 
ties,  which  are  marked  as  receiving  no  strain  in  the  end  panels. 


334  MECHANICS   OF   THE   GIRDER. 

It  is  therefore  usual  to  insert  these  members  in  the  end 
panels,  and  also,  in  this  style  of  girder,  to  stiffen  the  bottom 
chords  by  cross-bracing  in  the  end  panels,  so  that  each  bottom 
chord  may  there  act  as  a  strut. 

Some  builders  place  counters  in  all  panels,  even  where  the 
assumed  behavior  of  the  given  load  does  not  require  them.  By 
so  doing  they  provide  for  concentrated  loads  greater  than  the 
assumed  uniform  apex  load,  as  well  as  enhance  the  symmetry 
of  the  structure. 

In  short-span  bridges,  as  in  the  present  example,  some  of 
the  vertical  struts  require  a  greater  cross-section  than  the 
actual  downward  pressure  upon  them  would  indicate ;  for, 
besides  this  pressure  along  the  axis  of  the  strut,  it  should  be 
able  to  resist  probable  lateral  blows  from  the  travel  of  the  road, 
even  though  every  strut  be  protected  from  ordinary  collision 
with  hubs. 

To  provide  for  the  increase  of  bridge  weight  from  these 
sources,  above  the  weight  computed  from  the  given  load  nLy  we 
have  added  the  last  term,  F,  of  K,  which  term  should  be  large 
enough  to  cover  every  thing  not  otherwise  included. 

No  absolutely  definite  rule  can  be  given  for  the  size  of  these 
parts,  but  the  smallest  counter  ties  should  be  so  large  as  not  to 
look  wiry,  say  not  less  than  one  inch  in  cross-section ;  the 
bottom  chord  in  the  first  panel  may  equal  in  size  that  of  the 
second  panel ;  and  the  size  of  any  vertical  strut  should  enable 
it  to  resist  such  lateral  shocks  as  are  probable  in  the  situa- 
tion. 

132.  Thus  far  the  panel  weights,  W  and  L,  have  been 
assumed  to  be  applied  at  the  lower  joints  only.  In  the  nature 
of  the  case,  however,  it  is  plain  that  the  weight  of  the  top 
chords  and  the  system  of  head-bracing,  as  also  the  weight  of 
the  girder  diagonals,  can  only  reach  the  bottom  joints  through 
the  vertical  struts.  But  as  the  weight  of  these  members  is 
small  compared  with  the  whole  weight  of  the  bridge,  and  as 


GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      335 

the  calculation  is  a  little  more  simple  when  W  is  applied  at  one 
point  instead  of  two,  it  is  usual  to  make  the  above  assumption. 
It  is  proper,  however,  in  this  place,  to  indicate  the  changes 
to  be  made  in  the  strain  sheet,  Fig.  112,  by  changes  in  the  dis- 
tribution of  the  loads. 

ist,  Suppose  the  panel  weight,  L,  of  live  load,  and  the 
panel  weight  of  the  floor  system,  and  half  the  panel  weight  of 
the  girders,  to  be  applied  at  each  lower  joint,  and  the  other 
half  of  the  girders'  weight,  and  the  system  of  head-trussing,  to 
be  uniformly  distributed  at  the  upper  joints. 

We  have  G  =  weight  of  girders,  in  pounds. 

G  -T-  2.11  =  one-half  panel  weight  of  girders,  in  pounds. 

Take  A  =  panel  weight  of  head-bracing, 

X* 

/.     Load  at  each  lower  joint  =  L  +  W  —  A  --  , 

2H 

f 

Load  at  each  upper  joint  =  A  -\  --  pounds. 

2H 

The  strain  sheet,  Fig.  112,  applies  to  this  case  if  to  the 

Q 

compression  on  each  vertical  we  add  A  -\  —  -  pounds,  but  to 

each  end  post  \(  A  +  --Y     And  the  additional  weight  of  the 
verticals  due  to  this  change  of  loading  is 


^  V 
2nf 


which  is  to  be  added  to  the  weight  of  verticals  in  equation 
(426),  and,  of  course,  to  the  second  member  of  (429)  and  (430), 
and  thence  a  new  expression  for  G  be  found. 

2d,  Suppose  we  have  a  "deck"  bridge,  and  that  both  W 
and  L  are  applied  at  the  upper  joints. 


336  MECHANICS   OF   THE   GIRDER. 

Then  to  each  vertical  compression  given  in  strain  sheet, 
Fig.  112,  we  must  add  W  -\-  L  ;  and  the  additional  weight  of 
the  .verticals  is,  with  J(  W  -f-  L)  on  each  end  post, 

1™£L(W  4-  L)n  pounds, 

to  be  placed  in  the  second  member  of  (429)  and  (430),  provided 
the  bridge  has  its  points  of  support  at  the  bottom,  as  in  the 
figure ;  but  if  the  points  of  support  are  at  the  ends  of  the 
upper  chords,  then  no  end  posts  are  required,  and  their  weight 
may  be  deducted  from  the  second  member  of  (429)  and  (430), 
and 


be  added. 

3d,  In  case  of  the  deck  bridge,  if  we  suppose  half  the 
weight  of  the  girders,  and  the  weight,  nAlt  of  the  bottom  hori- 
zontal bracing,  to  be  applied  uniformly  at  the  bottom  joints, 
while  the  remainder  of  the  loading  is  applied  at  the  upper 
joints,  we  must  then  add  to  the  pressure  on  each  vertical, 
Fig.  112,  ,, 

L  -f  W  —  At pounds, 

2H 

instead  of  W  +  L.     And  the  additional  weight  of  girders  from 
this  source  is 


+  W  _  Ai  _        n          ds> 

. 

or 

W. 


_  Ai  _  <L\n  _  x)  pound 
2n/ 


minus  the  weight  of  the  end  posts,  according  as  the  girders  are 
supported  at  bottom  or  at  top. 


t 

GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      337 

133.  The  deck  bridge  requires,  especially  when  its  points  of 
support  are  at  the  bottom,  a  thorough  system  of  lateral  sway- 
bracing,  which  may  be  made  by  inserting  diagonals  between 
each  top  chord  and  the  bottom  chord  of  the  opposite  girder  at 
the  panel  joints,  in  addition  to  the  horizontal  systems  already- 
provided  for. 

The  proper  size  of  these  diagonals  can  be  determined  by 
calculation  when  the  applied  external  forces  are  given,  so  as  to 
conform  to  the  magnitude,  situation,  and  uses  of  the  structure. 

Their  weight  is  to  be  included  in  the  value  of  K,  the  con- 
stant part  of  the  bridge  weight. 

134.  To  determine  the  best  number  of  panels,  n,  and  the 
best  height,  //,  of  girder,  for  a  bridge  of  given  span,  /,  and  given 
moving  panel  load,  L,  we  may  find,  by  means  of  equations  (419), 
(429),   and   (430),   an  expression  for   W,  the  panel  weight  of 

bridge,  in  terms  of  n  and  h ;   then,  putting  ( — — )  =  o,  and 

\  an  / 

( — — -j  =•  o,  we  shall  have  two  simultaneous  equations  which 

will  yield  those  values  of  n  and  h  that  will  render  W  a  mini- 
mum. But  in  practice  it  will  be  found  more  convenient,  since 
n  is  always  an  integer,  and  the  two  simultaneous  equations  are 
of  a  high  degree,  to  find  W  in  terms  of  h  alone  for  several 
different  values  of  ;/,  presumably  including  the  best,  and  then 

to  find  from  — — -  ==  o,  for  each  value  of  n,  the  value  of  h  which 

a/i 

renders  W  least.  It  is  evident  that  the  values  of  n  and  h 
which  simultaneously  render  W  least  are  the  values  sought. 
For  the  present  purpose,  we  must,  of  course,  retain  ;/  and  h,  or 
their  equivalents,  wherever  they  occur  in  both  K  and  G.  Let 
us,  therefore,  re-examine  the  several  terms  of  K  and  G>  and  put 
them  into  suitable  form  for  general  application. 

The  value  of  F,  the  weight  of  floor,  (408),  is  independent  of 
;/  and  //,  and  requires  no  change. 


t 

338  MECHANICS  OF   THE   GIRDER. 

If  for  joists  we  call 

d  =  b\  (431) 

we  shall  have  a  good  ratio  of  breadth,  b,  to  depth,  d\  and,  in 
(410),  bd  =  fo,  and 


(432) 

2000;zZ)    '        (433) 


which  is  the  weight  of  the  joists,  in  pounds. 
Restoring  the  value  of  D,  we  write,  for  (414), 


JCF+   /+    2000«Z)  ?,/)!• 

/>=  15.460680*7,  (»-i)    -  —  ^g—  -f    pounds,    (434) 


'equal  to  weight  of  (n  —  i)  wrought-iron  I-beams  having  the 
proportions  assumed  in  deriving  equation  (412). 

135.  If  we  take  into  account  the  greatest  probable  pressure 
of  wind  horizontally  against  the  side  of  each  open  girder  and 
its  moving-load,  or  against  the  entire  side  of  each  wholly  cov- 
ered structure,  we  find  the  strains  due  to  wind,  in  the  chords 
and  entire  lateral  system,  by  making  the  proper  changes  in  the 
.strain  sheet,  Fig.  112. 

For  any  through  bridge  of  Class  IX.,  let  the  uniform  wind 
pressure  to  be  resisted  by  the  top  or  bottom  lateral  system  be 

Wl  =  \hw~  tons  per  panel  ;  w  being  the  horizontal  pressure 

of  wind  per  square  foot,  in  tons.  And  for  the  bottom  lateral 
system,  which  alone  is  affected  by  the  wind  pressure  against 
the  moving-load,  let  the  uniform  moving  wind  pressure  per 

panel  be  L,  =  sw~  tons  ;  c  being  the  height  of  train  or  other 
moving-load,  in  feet. 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      339 

From  (424)  we  derive  the  additional  weight  of  top  chords 
due  to  wind  pressure  by  substituting  2  Wt  =  -  -  for  ( W  -\-  L) ; 

since,  in  order  to  provide  for  the  wind  coming  either  way,  we 
must  increase  each  chord  for  increased  compression,  and  by 
putting  q  for  /i,  and  formulating  thus, 


Weight  of  top  chords)        mffihw,     2    .  v 

j  •    j  t  —       7T~^     V         T  on  —  2) 

due  to  wind          )         2  (V;z2^ 

(«  even), 

=  mffihw^  H 
2  Qn3g 

(n  odd). 


2  —  2n  —  3) 


flDjJ 


(435) 


Similarly,  from  (425),  putting  (2  Wt  -f  2L,)  =  —  (h  +  2s) 


for  ( W  +  L),  and  ^  for  //, 

Weight  of  bottom  )  _  mfl*w(h  +2g)  , 
chords  due  to  wind  J  2  Tr$q 

(n  even), 


x 
—  24) 


(»  odd). 


(436) 


And,  from  (426),  we  derive  the  weight  of  the  horizontal  struts 

between  the  top  chords  by  putting  Wt  =  \hw-  for  W,  o  for  Ly 

11 

q  for  //,  Q,  for  Qa  and  adding, 


by  reason  of  the  load  being  applied  to  the  compressed  chord, 
as  explained  in  article  132. 


340 


MECHANICS  OF   THE   GIRDER. 


Weight   of  top  horizontal )        ynfqwlh 
struts  due  to  wind       )  =        ^ 


=  U.      (437) 


The  floor  beams  which  carry  the  moving  load  generally  act 
as  the  horizontal  struts  between  the  loaded  chords ;  and  they 
are  usually  so  large,  in  comparison  with  the  struts  actually 
required  to  resist  the  wind  pressure,  that  we  may  with  little 
error  make  no  further  allowance  for  these  beams  acting  as  hori- 
zontal struts  than  that  already  suggested  in  article  124. 

But,  if  it  is  required,  we  can  find  the  additional  metal  to 
compensate  the  floor  beams  for  this  end  pressure  by  treating 
each  beam  as  a  pillar  whose  least  diameter  is  its  depth,  since 
the  longitudinal  joists  or  stringers  prevent  deflection  sideways. 

Thus,  ql  being  the  length,  d  the  depth,  of  the  wrought-iron 
I  floor  beams,  and  5  the  cross-section  due  to  the  total  effect  of 
wind  pressure,  P,  in  tons,  applied  longitudinally  at  the  end  of  a 
beam,  we  have,  from  equation  (400), 


square  inches, 


to  be  added  to  section  of  each  beam,  in  order  to  neutralize 
effect  of  wind  upon  the  loaded  horizontal  system  of  struts. 


-flP 


equals  total  additional  section  of  I-beams  ;  /  being  the  factor  of 
safety. 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      34! 

Now,  in  this  case,  2<P  takes  the  place  of  2^V  and  2<ZLy 
found  by  summing  the  vertical  strains,  Fig.  112,  and  used  in 
equation  (426),  provided  we  put  W^  for  W,  Lt  for  L.  For, 
adding  n(W^  +  ^i)>  since  the  load  is  applied  on  the  windward 
side  in  the  direction  of  the  wind's  motion,  and  subtracting  the 
pressures  then  upon  the  end  struts,  since  no  struts  or  I-beams 
are  used  on  the  abutments,  will  not  alter  ^P. 


Weight  of  I-beams  j    _ynfqJV*P 
due  to  wind,  from  >  Q3 

(426), 


(n  odd), 

where 


(433) 


.  18 

ad2 

/  :=  length,  in  feet,  between  centres  of  end  pins. 
ft  =  numerator  of  Gordon  formula  (400). 
n  =:  number  of  panels. 
u  =  constant.     (See  Table  IV.) 

h  =.  height  of  girders,  in  feet,  between  centres  of  chords. 
ql  =  entire  length  of  floor  beam,  in  feet. 
d  =  depth  of  beam,  in  inches. 

«  =  height  of  train  or  moving  wind-resisting  surface. 
w  =•  pressure  of  wind  per  square  foot,  in  tons. 

136.  In  finding  the  diagonals  of  the  horizontal  systems,  top 
and  bottom,  due  to  wind  pressure  applied  on  either  side,  we 
must  plainly  make  all  the  diagonals  mains,  and  the  two  in  any 
one  panel  each  equal  to  the  original  main  tie  in  that  panel. 

Using  the  strain  sheet,   Fig.  112,  as  a  horizontal  system 

now,  putting  W*  for  W,  Lt  for  Z,  q  for  //,  sin  <£,  =         n<?     — 

V/2  -|-  n*q* 

for  sin  </>,  Yl  for  Y,  the  strain  in  any  horizontal  diagonal  tie  due 


342  MECHANICS  OF   THE   GIRDER. 

to  wind,  we  have,  for  the  horizontal  system  between  the  loaded 
chords, 

Sum  of  horizontal  \        „  ,.        4  W\    (nz 
=  — 


m  of  horizontal  \        „  ,.        4  W\    (nz     /  n  \  )    1 

diagonal  strains      =  ^^  {—('+3+S+7  +  •  •  •  -  terms)  j 
due  to  wind 


+2  [i +3+6+IQ+  .  .  .  (|-i)  terms!  j 


W,n*  Z,      (     ,       j 

'2  sin  0i     12  sin  0! 


(n  even), 


(439) 


+  3  +  6  +  10 .  .  .  ^-3  terms^ 
+  the  (n~  l\    term  of  the  series  (i  +  3  +  6  +  10  ^ 


for  the  middle  panel  J 


(440) 


(n  odd). 

Therefore,  for  horizontal  system  uniting  loaded  chords, 

Weight  of  horizontal  1 

diagonals  due  to  \  =  SPi  X 
wind  pressure 


rsin2^ 


(n  even), 

'.  +  ?£,) 


(a  odd). 


(440 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      343 

137.  It  may  be  noted  here,  that,  however  complete  and  effi- 
cient the  horizontal  systems  are  made,  they  will  be  unable  to 
maintain  the  stability  of  the  bridge  under  the  action  of  wind 
if  the  posts  and  horizontal  struts  at  the  ends  of  the  bridge  are 
not  sufficient  to  resist  the  lateral  pressure  transmitted  to  them 
from  these  horizontal  systems.     That  is  to  say,  the  end  frame- 
work of   the  bridge  must  be,  with  regard  to  the  wind  force, 
incapable  of  lateral  motion,  whether  of  translation,  rotation,  or 
distortion. 

The  required  stability  may  be  secured  by  making  sufficiently 
large  end  posts  fast  to  the  abutments  for  light  and  high  struc- 
tures, and  by  attaching  these  end  posts  to  rigid  head  struts  by 
means  of  diagonal  braces.  But  as  all  this  excess  of  weight 
over  the  ordinary  panel  weight  rests  directly  upon  the  abut- 
ments, it  does  not  enter  into  the  formulae  for  strains  due  to  the 
uniform  panel  pressures,  W,  L  ;  Wlt  LT. 

This  excess  of  weight,  however,  has  an  influence  on  the  best 
values  of  n  and  h  ;  and,  calling  the  excess  Ew  pounds,  we  here 
proceed  to  formulate  its  value,  and  find  the  conditions  of 
stability. 

138.  To  find  the  additional  strains  and  weights  of  the  end 
members  of  a  bridge  of  two  girders  of  Class  IX.  required  to 
resist  a  given  wind  pressure,  let  Fig.  114  represent  the  elevation 
of  the  end  frame  of  a  through  bridge  of  this  class,  together  with 
its  full  moving-load. 

Then,  according  to  our  previous  notation,  the  total  hori- 
zontal pressure  at  A  is 


(442) 
and  at  B, 

P,  =  \n(W,+  A)  =  \wl(h  +  2e).  (443) 

The  vertical  pressure  on  each  abutment  is  \n(W  -f-  L). 


344 


MECHANICS  OF   THE   GIRDER. 


Now,  supposing  these  ends  of  iron  rest  upon  a  plane  stone 
surface,  and  calling  the  "co-efficient  of  friction"  for  iron  upon 
stone  |  (see  any  good  treatise  on  elementary  mechanics),  we 
must  have,  according  to  the  received  law  of  friction, 


Z), 


(444) 


which  is  the  condition  that  prevents  lateral  translation  along  a 
plane  stone  surface,  BE,  Fig.  114. 


For  stability  against  overturning,  — 


ist,  Without  live  load.     Take  the  moments  about  E\  we 
need,  since  AF  =  q,  and  FE  =  hy  as  above, 


(445) 


GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      345 

or,  if  each  end  of  the  girders  is  tied  to  the  abutment  with  a 
force,  /, 

qt,  (446) 


the  condition  that  prevents  rotation  of  unloaded  bridge  about 
the  points  of  support. 

2d,  With  live  load  resting  upon  a  beam  attached  to  the 
girders  at  the  ends  B  and  £,  we  require  the  condition 


(447) 

4 

girders  not  tied  down  ;  or 

t  (448) 


when  they  are  tied  with  the  force  /. 

But,  if  the  end  of  the  live  load  rests  directly  on  the  abut- 
ment, and  is  not  connected  with  the  girders,  the  condition  of 
stability  is 

P*h<  \q\Wn   +L(n-  i)],  (449) 

or 

P^h  <  Iq^Wn  +  L(n  -  i)]  +  q*.  (45°) 


3d,  For  the  stability  of  the  load  itself  against  turning  on 
its  own  points  of  support,  we  must  have 

P4e<L2g-,  (4501s 

s  being  the  height  of  moving-load,  g  the  gauge  or  breadth  of 
base,  and  P4  the  wind  pressure  acting  upon  the  part  L2  of  this 
load. 

The  strains  developed  in  the  frame  BAFE  will  be  greatest 
when  the  end  posts  cannot  move  laterally  at  the  bottom  nor 


346  MECHANICS  OF   THE   GIRDER. 

are  truly  fixed,  so  that,  when  under  wind  pressure,  the  tangents 
to  their  elastic  curves,  at  the  bases,  will  be  vertical. 

Let  b  —  length  of  brace,  GG»  in  feet,  and  (3  =  the  angle  it 
makes  with  the  vertical  post  ;  then  — 

Shearing-strain  at  any  cross-section  of  BC  or  GE  is 

s  =  \P*.  (452) 

Moment  at  any  point,  x,  above  B  or  E,  is  equal  to 

\P2x  =  M  =  \P2(h  -  ^cos/5),  at  C  or  G.      (453) 
Taking  moments  about  A  for  the  post, 
\P2h  =  Hb  cos  J3, 


which  is  the  horizontal  component  of  the  brace  strain  D. 

n  =  JL  =       p*h 

sin/3       2£sin/?cos/2 

in  tension  or  compression. 

Moment  at  any  point  between  Cs  and  GIt 

M  =  \P,h.  (456) 

>j  To  enable  each  end  post  to  resist  this  additional  moment, 
(453),  it  would  require  R  =  Bacd  as  the  moment  of  the  inter- 
nal stresses  on  the  added  material,  if  it  be  added  to  the  outsides 
of  the  post,  at  the  distance  \d,  in  inches,  from  the  neutral  axis  ; 
B  being  the  ultimate  bending  unit  strength  of  section,  a  =  width 
of  post,  in  inches,  and  c  =  the  uniform  thickness  of  additional 
iron  on  each  side  due  to  the  greatest  moment  at  C  or  G. 


GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      347 


Hence 


Whole  thickness  of  added  1  i2P2(h  —  />cos/3)/  . 

>    —  2C    —  —    --  *  - 

iron  j  aBd 

Cross-section    of  added  |  =  ^  =  ia/>(A  -  Jcosffl/  inches< 


ron 


Volume  of  4  posts   due  j  _  4  X  i22P2(/i  —  bco$(3)fh      ,  . 
to  wind  J  Bd 

Weight  to  be  added   to  ' 
4  posts  at  end  due 
wind  bending, 


Bd 


(457) 


Similarly,  from  (456),  for  the  two  top  horizontal  end  struts, 
of  length  q  feet,  and  depth  */,  inches, 


Weight  to  be  added  to  2  end  struts  to  1  __  2  x 

resist  bending  from  wind  force    J  ~  '  Bdl 

(458) 
pounds. 


If  ^2  is  the  least  diameter,  in  inches,  of  a  brace  of  length 
b  feet,  with  fixed  ends,  to  resist  the  longitudinal  pressure  D, 
(455),  then,  by  the  Gordon  formula,  (400),  we  have 


Cross-section  of  brace,  6"  = j- —  square  inches ; 

/,  as  before,  being  the  factor  of  safety,  and  fs  the  numerator  of 
Gordon  formula. 


/.    Volume  of  4  braces  =  48^  =  r cubic  inches. 


348  MECHANICS   OF   THE   GIRDER. 

Weight  of  4  braces  =  /.  sin  fi  cofy^ 

= -7—. — 5 5 — - —  pounds. 

/!  sin  p  cos  p 

\  h  — 


-- 

Bd  zBdt       24/x  sm  /3  cos  (3  j 


(46o) 


which  is  the  excess  of  weight,  in  pounds,  of  wrought-iron,  on 
the  two  abutments,  due  to  wind  pressure,  and  not  affecting  the 
uniform  panel  weight  of  bridge,  W. 

139.  In  the  preceding  investigation,  we  have  assumed  that 
the  entire  effort  of  the  wind  to  distort  the  rectangular  cross- 
section  of  the  bridge  is  to  be  resisted  by  the  two  end  frames 
alone. 

Instead  of  this  provision,  however,  we  may  fix  firmly  each 
horizontal  strut  of  the  unsupported  lateral  system  throughout 
the  bridge  to  the  ends  of  the  posts  abutting  upon  it,  and  thus 
transfer  the  whole  wind  pressure  to  that  lateral  system  which 
is  between  the  chords  resting  upon  the  supports.  The  same 
transfer  would  also  be  accomplished  should  we  connect  the 
posts  rigidly  to  the  other,  or  supported,  lateral  struts.  This 
procedure  would  enable  us  to  dispense  with  the  horizontal 
diagonals  of  the  unsupported  system  but  for  the  necessity  of 
retaining  them  to  keep  the  chords  they  connect  from  deflecting 
horizontally. 

In  this  case,  of  course,  the  horizontal  diagonals  and  struts 
of  the  supported  system  will  have  twice  as  great  horizontal 
pressure  to  resist  as  in  the  former  case,  and  (438)  and  (441) 
must  be  multiplied  by  2. 

The  horizontal  struts  of  the  unsupported  system  must  be 


GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      349 

able  to  resist,  in  a  vertical  direction,  the  bending-moment  found 
by  (456),  when  for  P2  we  put  Wiy  or  its  value  \wh-  ,  giving 


If  =  \wh-  =  (461) 

n          127 


for  each  strut  ;  d2  being  the  depth  of  horizontal  strut,  in  inches, 
a  the  width,  and  c  the  thickness  of  each  of  two  plates  of  iron 
added  at  the  distance  \d2  from  the  neutral  axis  of  the  strut. 
B  -T-  f  =  allowed  bending  unit  strain,  in  tons,  per  square  inch, 
since  w  is  in  tons.  Then  the  weight  of  all  these  horizontal 
struts  due  to  the  bending-moment,  (461),  must  be  the  same  as 
in  (458)  if  we  put  d2  for  dlt  and  regard  the  two  extreme  struts 
as  one,  since  each  sustains  but  half  a  panel  pressure. 

At  the  same  time,  these  horizontal  struts  must  resist,  in  the 
direction  of  their  least  diameters,  the  bending-moment  due  to 
the  longitudinal  strain  brought  upon  them  by  the  attached  hori- 
zontal diagonals  in  adjusting  the  bridge. 

Now  these  horizontal  diagonals,  between  unsupported  chords, 
may  be  of  uniform  size,  having  a  cross-section  S,  (say)  of  not 
less  than  about  I  square  inch.  Then,  if  the  allowed  unit  strain 
upon  them  is  T  4-  f,  and  if  their  inclination  to  the  plane  of 
the  girder  is  <£„  we  have  the  longitudinal  pressure  of  two  diag- 
onals, from  adjustment,  to  be  provided  for,  equal  to 


(46*) 


And  if  —  =  —  -  -  £-  -  -  =  the  allowed  pressure  per  square 


\ 
inch  upon  a  strut,  Q2,  Pa,  /„  and  T  being  of  the  same  denom- 


35O  MECHANICS   OF   THE    GIRDER. 

ination,  and  ft  =  numerator  of  Gordon  formula,  f  =  factor  of 
safety,  then 

„     ,    Q2  _  27IS'sin<ftr  _  ~  (   ,  v 

•^  -  —  7  --  75  -  —  "ii  (403) 

/  v:2 

which  is  the  cross-section  of  the  strut  due  to  the  end  pressure 

Pa- 

Hence,  from  (461)  and  (463),  — 

Total  section  of  a  horizontal  strut  between  the  unsupported 
chords  is,  in  square  inches, 

zl  .    2  TS  sin  0, 

+  '  -^'          (464) 


And  the  weight  of  n  horizontal  struts  between  the  unsupported 
chords,  to  resist  the  adjustment  and  distortion  strains,  is 


in  pounds,  where  n  is  used  instead  of  (n  +  i),  since  the  two 
extreme  struts  surfer  only  the  strain  due  to  any  one  of  the 
others. 

For  the  additional  iron  required  in  the  posts  to  resist  distor- 

tion by  the  wind,  we  have,  from  (453),  by  putting  -  -  for  P2y 
and  taking  the  moment  at  the  centre  of  post  where  x  =  \h> 


M  =  -   =  inch-tons  (466) 

2    X    4?2  / 

as  the  bending-moment  allowed  at  the  weakest  part  of  the  post, 
each  end  post  having  but  \M  instead  of  M.     Therefore 

Whole  thickness  of  added  iron  for  i  1  _        _  yvflh* 
post  anBd 


GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      351 

Cross-section   to   be   added    to   each  |  =  2^  =  3!^!          e  inches. 
post  j  nBd 


Weight  to  be  added  to  2n  posts  to 
resist    distortion    of    rectangular 


cross-section  of   bridge 


pounds.      (467) 


Finally,  the  weight  of  2n  wrought-iron  braces  for  this  case 
also  is  given  by  (459)  ;  and  the  cross-section  of  one  brace  is 


since  D  in  (455)  now  becomes 

wh2! 


^nb  sin  ft  cos  ft 

140.  We  will  now  exemplify  the  method  of  article  138,  which 
provides,  in  the  end  frames  alone,  the  means  of  resisting  the 
distorting  influence  of  the  wind. 

EXAMPLE.  — To  find  the  best  number  of  panels,  n,  and  the 
best  height,  7z,  for  the  two  wrought-iron  girders  of  a  highway 
"through"  bridge  of  100  feet  span  =  /,  and  1 8  feet  wide 
between  centres  of  chords  =  q,  single  system  of  Class  IX., 
Pratt  Truss,  under  a  uniform  rolling  load  of  I  ton  =  2,000 
pounds  per  running  foot,  in  addition  to  the  weight  of  bridge. 
Also,  to  find  the  weight,  n  W,  of  the  bridge  corresponding  to 
the  best  values  of  n  and  /i,  using  4  as  the  factor  of  safety  for 
iron,  and  10  for  wood,  and  taking  account  of  wind  pressure. 

Let  us  compute  for  n  =  5,  6,  7,  8,  9,  10,  ir,  12,  in  succes- 
sion, as  explained  in  article  134,  retaining  h  and  W  in  all  the 
expressions  for  weight. 


352  MECHANICS   OF   THE    GIRDER. 

ist,  The  floor  of  pine,  called  50  pounds  per  cubic  foot. 
Thickness  t  —  ^|  foot. 

Width  q'        =17.5  feet. 
Length  /       =  100  feet. 

2.5 
Weight  of  floor  F  —   —  x  17.5  X  100  x  50  =  18229  pounds. 

2d,  The  joists  of  pine  at  50  pounds  per  cubic  foot. 

g  =  2  feet  between  centres. 

B  •=.  7,000  pounds  per  square  inch  =  ultimate  resistance  to 

cross-breaking. 

/  =  10,  factor  of  safety  for  pine. 
/4-  n  =  panel  length  of  joist,  in  feet. 

d  =  b2  —  depth  of  joist,  in  inches,  by  (431). 
Then,  by  (432),  we  have 


Thickness  of)  X  10  x  2  x  100  H       .. 

200000)     «_-  ins.; 


a  joist,  *»'X  i7.5  X 
and,  from  (433), 

W    •    1,,      f   -    •   *         7  100    X     17.5     X     50/7.965443^  153548 

Weight  of  joists,  /  =          I44  x  2    -  V-^f—y  =  -*an  P°unds- 

3d,  The  wrought-iron  I-beams,  w  —  i  in  number,  supporting 

the  joists,  floor,  and  moving-load  L  =  -  =  —  tons  per  panel. 

;^         72 

Take  j5  =  50,000  pounds,  Table  II. 

Length  of  beam  q^  =  18.5  feet. 

K  1  5  3548  \i8.5  x 

- 


//   -f    2I8229V 

=  0.4331885^-    —  -  --  J  inches, 


from  (412),  using  the  proportions  assumed  in  finding  that  equa- 
tion. 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      353 

By  (434)> 

Weight  of  I-beams,  P  =  15.  46068  (n  —  i)  X  -^ 

1  8 

l8-     X 


50000 


//  +  218229X1 
=  i.  03  1  824  (n  —  i)(  -  -  -  )  pounds.. 


4th,  The  horizontal  struts  of  the  top  lateral  system  of  this 
"through  "  bridge. 

In  this  example  of  a  highway  bridge,  let  us  assume,  as  actual 
pressure  of  wind  per  square  foot,  the  large  value  75  pounds ; 
also  that  the  two  open  girders  offer  a  resisting  surface  equiva- 
lent to  |  of  the  surface  presented  if  the  bridge  were  covered, 
that  is,  equal  to  f ///.  Then  the  whole  wind  force  to  be  resisted 
is  |  X  75///  =  45/*/  pounds. 

Wind  force  per  running  foot  —  45/2  pounds. 

Wind  force  per  square  foot     =  w  =  2$$$  =•  0.0225  ton. 

Although  this  wind  force  is  actually  applied  to  both  girders, 
we  shall  regard  it  as  distributed  equally  to  the  panel  points  of 
the  two  windward  chords,  no  account  being  here  taken  of  the 
action  of  wind  on  passing  carriages. 

Suppose  the  top  horizontal  struts  to  be  I-beams,  the  square 
of  whose  least  radius  of  gyration  is  r2  =  0.5  inch,  which  cor- 
responds to  a  six-inch  beam  of  ordinary  make.  Then,  using 
equation  (385),  and  calling  C  =  40,000,  E  =  27,300,000,  we 
have,  in  (437), 

20 

Q2  = =  4. 680056  tons; 

40000  x  2 io2 

47T2  X   27300000  X  0.5 


354  MECHANICS   OF   THE   GIRDER. 


and  (437)  gives 

Weight  of  top 
horizontal 
struts  due 
to  wind 


0.0225  x  iQohfn 


18  x  4.680056 


=  28.84581^-  +  2^  (n  even), 

=  28.8458I//7*2  ~  *  +  2^  («  odd). 


5th,  The  horizontal  diagonals,  top  and  bottom.  From  (441), 
where  now  Ll  —  o,  since  we  take  no  account  here  of  wind 
against  live  load  on  this  highway  bridge,  we  have,  making 
T  =  24  tons,  q  =  1  8  feet,  m  =  -f$  for  wrought-iron  (as  above), 


Weight   of  horizon-  } 

tal  diagonals,  top     =  X=  2  X  5  X  4  X  18  X  6         ^  («  even) 
and  bottom,         J  18X24  sin'*,          i  («z-  i)(«odd), 

where  i      —  ^  -L.  JL  =  i  +  IOQOQ 

sin2.^  ~  n2q2  ~         ~   iS2n2' 

6th,  Let  the  residual  weight,  Yy  be  1,000  pounds  for  all  val- 
ues of  n. 

7th,  The  additional  weight  of  iron  needed  in  the  I-beams, 
by  reason  of  their  acting  as  horizontal  struts  for  the  wind 
pressure  on  lower  chord,  is  found  by  (438),  after  computing  d 

as  already  formulated  for  the  floor  beams.    Here  Wt  =  -  tons  ; 

211 

LI  —  O]  /x  =  1  8  tons;  ql  =  18.5  feet;  a  =  750,  since  the  ends 

,0 

are  not  fixed.     Hence,  from  (438),  where  Q3  =  -  —, 

i  J-   2222 


GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      355 


Weightto  beadded  | 
to  floor  beams  \ 
due  to  wind 


3  x  5  x  4  x  l8.5  x  O.O225  x 


~ 


75°^ 


i8x  i8x  2 


«  (n  even), 


3.85416667/1  + 
\ 


X  n  (n  even), 
x^!nJ  (»0dd) 


Collecting  the  terms  of  K  thus  found,  we  have,  in  terms 
of  //,  — 

WEIGHTS  OF  THE  COMPONENTS  OF  K,  IN  POUNDS. 


« 

5 

6 

1 

8 

Floor     .     .     . 

18229.0000 

18229.0000 

18229.0000 

18229.0000 

Joists     .     .     . 

22258.0000 

17884.0000 

14864.0000 

12663.0000 

(  I-Beams      .     . 

5459.0000 

5969.0000 

6408.0000 

6796.0000 

I  Do.  wind    .     . 

23.39847* 

30.12457* 

35-37QI/& 

42^30867* 

Hor.  struts 

1  26.92  1  6/1 

144.22917* 

156.59147* 

173.07497* 

Hor.  diags.     . 

120.6662/1 

125.37047* 

125.7336^ 

13340257* 

Residual     .    . 

IOOO.OOOO 

IOOO.OOOO 

IOOO.OOOO 

IOOO.OOOO 

K   .     .     \ 

46946.0000 

43082.0000 

40501.0000 

38688.0000 

} 

+270.9862/6 

+  299.7  240/fc 

+317.69517* 

+348.78607* 

n 

9 

10 

11 

12 

Floor     .     .    . 

18229.0000 

18229.0000 

18229.0000 

18229.0000 

Joists     .     .     . 

10994.0000 

9688.0000 

8641.0000 

7784.0000 

(  I-Beams      .     . 

7146.0000 

7465.0000 

7760.0000 

8035.0000 

•  Do.  wind    .     . 

48.118  i/i 

55-33  12/' 

61.62267* 

69.1289/5 

Hor.  struts 

185.89507* 

201.9207/6 

215.02947* 

230.76657* 

Hor.  diags. 

138.10407* 

147.22207* 

154.03257* 

i63-9358/' 

Residual     .    . 

IOOO.OOOO 

IOOO.OOOO 

IOOO.OOOO 

'  IOOO.OOOO 

*•             \ 

37369.0000 

36382.0000 

35630.0000 

35048.0000 

\ 

+372.11717* 

+404.47397* 

+430.6845/5 

+463.83127* 

356 


MECHANICS  OF   THE   GIRDER. 


8th,  The  top  chords,  of  2  channels  and  2  plates  of  wrought- 
iron. 

In  each  panel  let  the  ratio  of  chord's  length  to  least  diam- 
eter be  15. 

Then,  in  (424), 

18 


=  16.7442  tons, 


+ 


by  (400). 


3000 


Weight  of  top  chords  due  to  ver- )  _       5  X  4  X  ioo2      /  yp  ,   ioo\ 
tical  pressures,  in  pounds,      j      2  x  18  x  1 6.7442^  \  n  ) 

X  2"2  +  3«  -  a  (n  even), 


~  3  (n  odd). 


And,  from  (435), 

Weight  of  top  chords  due )  _  5  x  4  X  ioo3  X  0.0225^ 
to  wind,  in  pounds,      j       2  x  18  x  16.7442  x  18 


-  2«  -  3  (w  odd). 


9th,  The  bottom  chords,  of  flat  links  or  I-bars. 
From  (425), 

Weight  of  bottom  chords  due  to )  _  5  x  4  X  ioo2/ ^  +  ioo\ 
vertical  forces,  in  pounds,       j        2  x  18  x  2^\  n  ) 


—  3«2  4- 


-  24 


—    21 


(n  even), 
(«odd). 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      357 


And,  from  (436),  s  being  zero, 

Weight  of  bottom  chords )        5  X  4  X  TOO*  x  0.0225^ 
due  wind,  in  pounds,  )  "          2  X  18  X  24  X  18 

2^3  _  3^2  +  22n  —  24 


«3 
-    3«2    4-    22H   -    21 


(n  even), 
(«odd). 


i  oth,  The  verticals.     Take  ratio  of  length  to  least  diameter 
30;  then,  in  (426), 


-  8.8 


750 


if  the  ends  are  not  fixed,  and  we  have 

Weight  of  verticals,  in  pounds, 

*  4-     5  X  4  X  iQQ/fr   hn*  +  3^  -  i<A 
2  X   18  X  8.i8i8\  n  ) 


_   3  X  5  X  4 
18  X  8.1818 


(n  even), 


=   3  X  5  X  4 
18  x  8.1818 


Q  I 

' 


x  4  X 


2  X  18  X  8.i8i8\ 
(n  odd). 


3\ 
J 


nth,  The  girder  diagonals,  by  (428). 

_  /* 

' 


sn2 


Weight  of  girder 
diagonals,  in  [•  = 
pounds, 


3X5X4^ 


Wn 


n 
(n  even), 


X4X  ioohfn2  —  i\         3X5  X. 
:24sin2</>  \    n    )       18x245! 


18x24  sin 


4^ 


(«  odd). 


358 


MECHANICS  OF   THE   GIRDER. 


Computing  for  the  different  values  of   ;z,   collecting,  and 
arranging,  we  have,  including  the  values  of  K  above,  — 


WEIGHTS  IN  POUNDS,     W  IN  TONS,    h  IN  FEET. 

Live  load  =  nL  =  100  tons,    I  =  100  feet. 


11 
5 

Top  chords     .}££• 

Bo,«om  chords  |£-. 
Verticals      

4140-742 
2444.444 

W 

h 

9.77778 

Wh 

82815 
48889 

i 
k 

- 

h* 

103.5185 

61.1111 

208.5926 

88  8889 

K  .    . 

46946 

270.9862 

2ooonlf  = 

7918.519 

+13.11111 

+167259 

+46946 

+733-0973 

6 

Top  chords     .J5£- 

Bottom  chords  {  Load  • 
I  Wind  . 

Verticals 

4866.256 
2777.778 

14  66667 

81104 
46296 

- 

101.3803 
57,8704 

1388  88q 

K  .    . 

43082 

299.7240 

2OOO«  W  = 

9032  923 

+19.66667 

+157407 

+43082 

+861.2381 

7 

Top  chords     .{£-• 

Bottom  chords  j  Load  • 
'  Wind  . 

Verticals 

S52S-323 
3174.604 

78933 
45351 

- 

98.6665 
56.6894 

6  66666 

K  .    . 

40501 

317-6951 

zooonW  = 

10060.471 

+26.22222 

+150199 

+40501 

+906.0064 

8 

Top  chords     .}£-• 

Bottom  chords  1  Load  ' 
'  Wind  . 

Verticals           .... 

6221.067 
3559-027 

26  07408 

77763 
44488 

- 

97.2042 

55-6098 

392.1296 

Diagonals   
K  .    . 

1388.889 

8.88889 

22787 

38688 

I45-8333 
348.7860 

2ooon  W  — 

11168.983 

+34.96297 

+145038 

+38688 

+1039.5629 

GIRDERS   WITH  EQUAL   AND   PARALLEL    CHORDS.      359 


WEIGHTS  IN  POUNDS,     W  IN  TONS,    h  IN  FEET. 

Live  Load  =  nL  =  100  tons,    /  =  100  feet. 


It 
9 

Bottom  chords  j  Load  ' 
,              f  Wind  . 

Verticals      

6881.576 
3978.051 

W 

k 

32.59260 

Wh 

76462 
44201 

i 

- 

r 

95-5774 

55-2507 
402  3777 

ii.  mil 

20322 

K  .    . 

37369 

372.1171 

2OOOH  W  = 

12231.369 

+43.70371 

+140985 

+37369 

+1089  9319 

10 

Top  chords     .  I  Load  ' 
'  Wind  . 

Bottom  chords  j  Load  • 
'  Wind  . 

7564.819 

4388.889 

75648 
43889 

- 

94-5602 

54.8611 
488  8889 

Diagonals    

1388.889 

13.88889 

18 

K  .    . 

- 

36382 

404.4739 

20oonJV  = 

13342.597 

+54.62964 

+137870 

+  36382 

+1226.1174 

ii 

Top  chords     .  j  Load  ' 

Bottom  chords  \  Load  ' 
f  Wind  . 

Verticals 

8226.202 
4820.936 

48  88888 

74784 
43827 

- 

93.4796 
547834 

1377.410 

1  6  66667 

16696 

35630 

430.6845 

2OOO«  W  — 

14424.548 

+65-55555 

+135307 

+3563° 

+1279.2841 

12 

Top  chords     .jLoad. 
'  Wind  . 

Bottom  chords  j  Load  ' 
'  Wind  . 

Verticals 

8903.037 
5246.912 

eR  66667 

74192 
43724 

- 

92.7400 
54.6553 

Diagonals 

1388  889 

K  .    . 

35048 

463-8312 

2000^  = 

15538.838 

+  78.66667 

+133241 

+35048 

+1416.9878 

-°-3959259  +  o-S*  ~ 

7-87035  -f  2.1541^  +  0.04306191^  , 
—0.4516461  -f  0.6/1  —  0.000983333^ 

7-5°995  +  2.02505^  +  0.04530032^ 


360  MECHANICS  OF  THE   GIRDER. 

Multiplying  each  of  these  eight  equations  by  — - — ,  we  find 

20000 

the  uniform  panel  weight  of  bridge,  W,  in  terms  of  // ;  thus  : 

8.36295  4-  2.3473^  4-  0.03665487/^2 

n  =    5,     W  = 

n  =    6,     W  = 

n  =  8,  W  = 

n  =  9,  W  = 

n  =  10,  W  = 

n  =  n,  W  = 

/Z    ?*"""    1 2         t^ff   —     . 

—  0.7769419    4-    1.2/1    —    0.00393333/^2 

In  differentiating  these  and  similar  expressions  for  W,  it 
will  be  convenient  to  have  a  typical  form  or  mode  of  operation. 
Let 

W  =    a  +  bh  +  ^  (469) 

be  a  type  of  these  equations  ;  then,  after  putting  — -  =  o,  and 

all 

reducing,  we  have  the  equation 

'•J>  4-  (acl  —  atc)(2h)  +  (be*  —  b^c)h*t      (47°) 


—  0.50302355  -f  0.7^  —  o. 

7.2519  +  1.9344^  +  0.05197814^ 
—0.55844915  +  0.8/1 

7.04925  +  1.86845^ 
—0.6115684  +  0.9^  —  0.002185185^ 

6-8935  +  1.8191^  4-  0.06130587^ 

—  0.6671298  +  h  —  O.OO273I482/Z2 

4-  1.7815^  +  0.06396421^ 


—0.7212274  +  i.iA  —  0.0032777777^ 
6.66205  +  1.7524^  4-  o. 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      361 


from  which  h  is  easily  found  :  and  there  is  no  need,  in  these 
cases,  of  taking  the  second  differential  to  ascertain  whether  the 
positive  value  of  k,  to  be  found  from  (470),  renders  W  a  maxi- 
mum or  a  minimum  ;  for  the  substitution  of  a  member  a  little 
less  or  a  little  greater  than  the  positive  value  of  h  so  found 
will  at  once  serve  to  verify  the  work,  and  show  W  to  be  a  mini- 
mum in  (469)  when  h  takes  the  value  given  by  (470). 

Taking  the  case  where  n  =  9,  and  using  logarithms,  we 
may  solve  thus  :  — 


n  =  9. 

Logs. 

Log  Co-efficients. 

Co-efficients. 

Equation  (470). 

a  =    7.049250000 

0.8481429 

logabi  =  0.8023854 

abl         6.3443300 

_ 

a-i  —  —0.611568400 

9.7864451 

Iogai3  =  0.0579266 

—a-^b          1.1426900 

7.4870200 

b    =     1.868450000 

0.2714815 

log&Ti  =7.6109697 

bc\       —0.0040829 

- 

bi  =     0.900000000 

9.9542425 

logfiic  =  8.6906118 

—  b\c       —0.0490469 

=  0.0531298/42 

c    =    0.054496590 

8.7363693 

log  ccii  =  8.5228144 

—  ca.1      +0.0333284 

- 

c\  =  —0.002185185 

7.3394882 

\ogcia  —  8.1876311 

da      —0.0154039 

—  0.0179245  (zh) 

0.0531  298/5* 
8.7253382 

—  0.0179245(2/0 
8.2534471 
9.5281089 
-  0.33737(2/5) 
9.0562178 
2.1493213  • 
1.0746607 

=     7.4870200 
0.8743090 
2.1489708 
=    140.9194000 
=  log  +0.1138000 
=  log  141.0332000 
=  log  ±11.  87  57400 

+0.3373700 

log  co-efficients, 
log  quotients, 

2  log  0.33737, 

log  (141.0332)* 
£  co-efficient  of 


When  W  is  a  minimum,  h  =.  12.21311  feet. 


log/5, 

1.0868263 

a 

zr 

7.049250 

log  6/1, 

1.3583078 

bh 

rr 

22.819590 

log//2, 

2.1736526 

log  chz, 

0.9100219 

chz 

= 

8.128720   37.99756o   I-S797SS7  =  log  num. 

log  c  x//2, 

9.5131408 

ctf 

= 

-  0.325942 

<zx  =  —  0.611568 
b-Ji  —  0.9/1  —      10.991799    10.054289   1.0023513  =  logdenom. 

nW  —  34.013151  tons.     W—    3.7792390.5774044  = 


362 


MECHANICS  OF   THE   GIRDER. 


Computing  h  and  Wn  for  the  other  values  of  n,  we  find 
them,  when  Wn  is  a  minimum,  as  follows  :  — 


Span  =  /  =  100  Feet.    Uniform  Live  Load  =  nL  =  100  Tons. 


No.  of 
Panels, 
n. 

Panel 
Length, 
/  -f  »  feet. 

Best   Height 
in  Feet, 
*, 

Ratio  of 
Length  to 
Height, 

Inclination  of 
Diagonals 
to  Horizon, 
* 

Minimum 
Bridge  Weight, 
wJFTons. 

Ratio  of 
Dead 
to   Live 

Load. 

Ratio  of 
Dead 
to  Total 
Load. 

5 

20 

16.50041 

6.0604 

39°  31"  26" 

37.177990 

0.37178 

0.27102 

6 

i6§ 

14.71481 

6-7959 

41°  26'  30" 

35.930016 

o-3593o 

0.26433 

7 
8 

147- 
rai 

I3-89555 
12.74032 

7.1966 
7.8491 

44°  12'  24" 
45°  32'  44" 

34.642979 

34.509872 

0.34643 

0.34510 

0.25730 
0.25656 

9 

»*i 

12.21311 

8.1879 

47°  42'  18" 

34-013151 

0.34013 

0.25380 

10 

10 

11.39809 

8-7734 

48°  44'  1  8" 

34.302300 

0.34302 

0.25541 

ii 

9TT 

11.02062 

9.0739 

50°  28'  51" 

34.156870 

°-34T57 

0.25460 

12 

8i 

10.40797 

9.6080 

51°  19'    i" 

34.635012 

0.34635 

0.25725 

Hence  34.013151  tons  is  the  least  of  these  least  weights,  or 
the  minimum  minimorum. 

By  observing  the  first  differences  of  the  values  of  n  W,  we 
may  perceive,  that,  in  addition  to  the  fact  that  the  odd  number 
»  =  9  gives  the  lowest  value  of  n  W,  the  odd  number  n  =  1 1 
gives  a  lower  value  of';/  W  than  either  of  the  even  numbers,  10 
and  12,  adjacent  to  it,  and  that  n  =  7  renders  nW  nearly  as 
small  as  n  =  8.  We  may,  therefore,  almost  infer  from  these 
eight  cases,  that,  when  near  the  best  value  of  ;/,  an  odd  number 
of  panels  is  preferable  to  an  even  number.  And  this  conclusion 
harmonizes  with  the  fact,  that,  in  case  of  an  odd  number  of 
panels,  there  is  no  weight  applied  at  the  centre  of  span  as  there 
is  when  n  is  even.  We  may  further  observe  that  the  difference 
between  the  greatest  and  least  values  of  n  W  in  these  eight 
cases  is  only  3.164839  tons,  provided  the  best  values  of  h  are 
used ;  but,  if  other  values  of  h  are  employed,  n  W  departs  more 
widely  from  its  least  value. 


GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      363 

Also  in  the  present  case,  when  n  W  is  least,  the  inclination, 
<f>,  of  the  girder  diagonals  to  the  horizon  is  about  2|  degrees 
above  45  degrees  ;  and  from  this  point  </>  increases  or  decreases 
with  11  if  the  best  value  is  given  to  h. 

The  best  ratio  of  length  to  height  of  girder  for  this  span 
and  load  is  8.1879;  and,  near  the  best  simultaneous  values  of 
n  and  /z,  we  have  approximately 

h  =  '-.  (47i) 

1 2th,  Let  us  now  find  the  value  of  Ew,  equation  (460),  the 
quantity  of  wrought-iron  to  be  added  to  the  end  framework  to 
resist  wind  force  tending  to  produce  distortion,  assuming  that 
the  bridge  is  so  fixed  to  the  abutments  that  neither  sliding  nor 
overturning  can  take  place. 

In  equation  (460),  take  b  =  4  feet  =  length  of  brace, 
d^  =z  6  inches,  /3  =  45  degrees  =  inclination  of  brace  to  post. 

/.     sin  ft  =  cos/3  =  0.70711,     <5cos/?  =  2.82844  feet. 

Take  d  =  12  inches,  width  of  end  post  to  resist  bending. 
di  =  12  inches,  depth  of  end  horizontal  strut. 
f,  =  1 8  tons. 
B  =  25  tons. 
q  —  1 8  feet. 
m  =  -fg  pound. 

/  =  4- 

/  —  100. 
w  =  0.0225  ton. 

Then,  computing  Ew  for  the  eight  values  of  h  already  found, 
we  obtain,  from  (460)  and  from  the  table  just  given,  the  follow- 
ing results :  — 


364 


MECHANICS  OF   THE    GIRDER. 


W  in  Tons,     h  in  Feet. 


No.  of  Panels,  n. 

5 

6 

7 

8 

Height,  h     

16.500 

14.71  S 

1-5  806 

12  74O 

n  times  panel  weight,  nW     .     . 
Added  iron,  Ew  tons     .... 
Weight  of  bridge,  n  W  +  Ew   . 

Weight  of  wood  

37.178 

3-935 
41.113 

2O  24.4 

35-927 
2.898 
38.825 

l8.(X7 

34.643 
2.489 

37-I32 

1  6  C47 

34-510 
1.980 
36.490 

I  ^.4.46 

20869 

20.768 

"o.  ;8<; 

2  1  .044 

Cost  of  iron,  at  $150     .    .    .    . 
Cost  of  wood,  at  $15     .    .     .     . 

$313°  35 
303  66 
7474  01 

$3115    20 

270  86 
3386  06 

$3087  75 

248    21 
-5-J7C    06 

$3156   60 
231    69 

-n88  20 

Q8  (X 

so  10 

^2    77 

No.  of  Panels,.  n. 

9 

10 

11 

12 

Height  h 

12  21  7 

1  1.  70.8 

II  O2I 

10.408 

n  times  panel  weight,  n  W     .     . 
Added  iron,  Ew  tons     .... 
Weight  of  bridge,  n  W  +  Ew    . 

34.013 

1773 
35-786 

I4.6l2 

34.302 
1.480 
35782 

I7.QCQ 

34-157 
1-357 
35-5I4 

I"V4T? 

34.635 
1.170 

35-805 
13.007 

^Veight  of  iron          ... 

21  174 

21  823 

22.O7Q 

22.708 

Cost  of  iron,  at  $150     .    .     .    . 
Cost  of  wood,  at  $15     .    .    .     . 
Cost  of  bridge 

$3176  10 
219  18 

•3-JQC     28 

$3273  45 
209  39 
3482  84 

$33"    85 

201  53 

•5  C  I  'J     ~l8 

^3419  70 
T95  ii 
7614  81 

Excess  over  least      .... 

Jjyj  f* 
CQ   -?2 

146  88 

177   42 

278  8q 

Here  we  see  that  n  =  u  and/2  =  11.021  are  the  conditions 
yielding  least  total  weight  of  bridge,  while  the  whole  cost  is  a 
minimum  if  n  =  7,  h  =  13.896,  and  (/  -f-  n)  =  14^. 

Notice  that  both  of  these  minima  of  weight  and  cost  corre- 
spond to  an  odd  number  of  panels,  and  that  the  excess  of  cost 
above  the  lowest  would  in  all  cases  more  than  compensate  the 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      365 

manufacturer  for  having  the  best  simultaneous  values  of  n  and 
h  determined  by  calculation,  as  above. 

If,  however,  there  would  be  sufficient  head-room,  we  may,  for 
this  span  and  load,  adopt  either  8  or  9  panels,  giving  more  iron, 
less  wood,  and  less  total  weight,  with  a  small  increase  of  cost. 
In  each  of  these  8  cases  it  will  be  seen  the  bridge  weight 
is  a  little  more  than  one-third  the  uniform  moving-load,  100 
tons  =  nL,  and  that  the  total  dead  load  is  slightly  greater 
than  one-fourth  the  sum  of  dead  and  live  loads. 

141.  To  exemplify  the  Method  of  Article  139,  which  pro- 
vides, at  Every  Post,  the  Means  of  resisting  the  Distorting 
Influence  of  the  Wind.  —  Taking  the  example  of  article  140, 
and  calling  the  top  horizontal  diagonals  I  inch  in  diameter  (that 
is,  0.7854  square  inch  cross-section),  and  weighing  2.654  pounds 
to  the  foot,  we  have 


Weight  of  2n  horizontal  top)  =  654V/i8*  +  ™* 

diagonals  )  "V  n* 


10000  pounds, 

Weight  of  bottom  horizontal)  __  y  (as  found  in  article  140,  for  both 
diagonals  )          L>  (        top  and  bottom. 

Strain  on  a  top  horizontal  strut,  from  ^  =  6  tons  per  square 
inch  on  two  top  diagonals,  is  equal  to 

2  x  6  x  0.7854  sin  0!  =  9.4248  sin  fa  tons. 

Now  we  already  have  the  breaking  inch  strain  on  top  hori- 
zontal struts  ==  4.680056  tons,  and 


366  MECHANICS  OF   THE   GIRDER. 

Therefore,  in  square  inches, 

Cross-section    of    a  \ 

top  strut  to  re-  I          9.4248      .  I          i 

sist  initial  strain  =  1.170014  sin*'  =    8'°553   /  ^       IOOOQ"  S*' 

on  diagonals        J  V  324722 

And,  from  (461), 

Cross-section    of    a    top  1  ,     /.,,, 

6wf//i2  6  h2  .     , 

strut  to  resist  distort-  \  =  2ac  =    -=*—  =  0.30-  -  square  inches 

-  r    •   j  Bd2n  7  n 

mg  force  of  wind          j 

if  w  =  0.0225  ton,  f  =  4,  /  =  100,  B  =  25  tons,  and  d2  =  7 
inches. 

From  (465),  since  \2mqn  —  12  X  y\  X  i8w  = 


Weight  of  »  top  hori-  J  +  ^^fc^t 1 unds 

zontal  struts          )  V  n2  -f-  30.8642 

/&  /        ~~i 

=  i85i.428-+48^.3i8^2l/ pounds, 

?2  V  /z2-}-  30.8642 

approximately,  by  reason  of  (471),  to  avoid  the  second  power  of 
h,  for  convenience. 

Weight  to  be  added  to  floor  beams  due  to  wind  =  two  times  Pf, 

as  already  given. 

In  (467)  take  d  ==  8  inches ;  then 

Weight  to  be  added  to  all  posts)        i22  x  5  X  4  X  0.0225  x  100 
to  resist  distortion  J  ~  2  x   18  X  25  x  8 


£22*  pounds, 


by  (471). 


GIRDERS   WITH  EQUAL   AMD  PARALLEL    CHORDS.      367 


In  the  previous  case  the  quantity  of  iron  of  uniform  thick- 
ness to  be  added  to  each  post  is  that  due  to  the  greatest  mo- 
ment given  by  equation  (453).  It  is  plain  from  that  equation 
that  the  added  iron  may  vary  in  thickness  from  C,  Fig.  114, 
where  it  should  be  greatest,  to  the  bottom,  where  it  may  be 
nothing.  Or,  without  increasing  the  thickness  of  the  iron,  the 
post  may  be  made  broader  at  top  than  at  bottom,  and  thus 
resist  the  bending-moment  whenever  this  broadening  is  not 
accompanied  by  too  great  reduction  of  the  thickness  of  the 
iron  composing  the  post.  In  the  present  case  x  =  \h. 

Finally,  from  (459),  calling  d2  =  4  inches, 


Weight  of  all)        6  X  5   X  4  X  0.0225   x   ioo 
braces      \  18  x  18  X  o.yoyn2 

=  i.  746  6  7  Aa 

=  174.667-  pounds, 
n 

by  (471). 

Computing  for  8  values  of  n,  we  find,  — 


Weights  of  the  Components  of  K,  in  Pounds. 


^— V 

>o  X  42/ 


3000  x 


n. 

5 

6 

7 

8 

Floor  

18229  oooo 

18229  oooo 

18220  oooo 

l82''9  OOOO 

Joists      

22258  oooo 

17884  oooo 

14864  oooo 

12667  oooo 

(  I  floor  beams  .... 
(  Do.  wind     

5459.0000 

46  7968/2 

5969.0000 

60  2490^ 

6408.0000 

7O  7AO'J/i 

6796.0000 

84  6172^ 

Horizontal  top  struts    < 
Horizontal  diagonals    < 

1617.0000 
370.2857/2 
714.0000 
120.6662/5 

T.A.QT.T.'ik 

2128.0000 

308.5714^ 

781.0000 

125.3704* 

29.1111^ 

2650.0000 

264.4898/5 
854.0000 

125.7336/5 

24.  Q?24/J 

3176.0000 
231.4286/2 
931.0000 
1334035* 

21.8777^ 

IOOO.OOOO 

JOOO.OOOO 

I  OOO.OOOO 

IOOO.OOOO 

'{ 

572.6820/& 
+  49277 

523.3019/2 
+45991 

485.9160/5 
+  44005 

47I.28l6/2 
+  42795 

368 


MECHANICS   OF   THE    GIRDER. 


Weights  of  the  Components  of  K,  in  Pounds.  —  Concluded. 


n. 

9 

10 

11 

12 

Floor  

18229.0000 

18229.0000 

18229.0000 

18229.0000 

Joists  

10994.0000 

9688.0000 

8641.0000 

7784.0000 

j  I  floor  beams  .... 
(  Do  wind         .     .     . 

7146.0000 
06.2^62^; 

7465.0000 
110.6624^ 

7760.0000 

1  2^.24.^2/1 

8035.0000 
n8.  2^78/2 

Horizontal  top  struts    < 
Horizontal  diagonals    < 

3701.0000 
205.7143,5 
IOII.OOOO 

138.1040/6 
19.4074/2 

4225.0000 
185.1429/2 
1093.0000 
147.2220/2 
17.4667/2 

4746.0000 
168.3117/2 
1177.0000 

154.0325/2 
15.8788/5 

5263.0000 

154.2857/5 

1263.0000 

1  63-93  58/' 

14.5555/2 

Residual          .              . 

IOOO.OOOO 

IOOO.OOOO 

IOOO.OOOO 

IOOO.OOOO 

*;{ 

459.4619/5 

+42081 

460.4940/2 

+41700 

461.4682/5 

+41553 

471.0348/2 
+41574 

The  strain  throughout  each  top  chord  due  to  the  initial 
strain,  6  X  0.7854  =  4.7124  tons  on  each  diagonal  between 

top  chords,  is 

4.  7  1  24  cos  0!  tons, 

and  the  allowed  inch  strain  here  is 

16.7442 

-  =  4.18605  tons.  , 

Therefore  the  additional  cross-section  of  iron  for  both  top 
chords  due  to  initial  strain  on  top  diagonals  is,  in  square 
inches, 

225.148  - 


4.18605 

Additional  weight  1 
for  top  chords, 
pounds,  due 
initial  strain  on 
top  diagonals 


=  2.25148  cos  fr  = 


1 
Y324722  +-  10000 


_  12  x  IPO  X  5  X  225.148 


75°49-333 


ioo 


¥324^  +-  i  oooo 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.       369 

The  effect  of  wind  on  the  bottom  chords  in  this'  case  will 
be  twice  what  it  was  in  the  example  of  article  140,  and  may  be 
taken  from  the  table  therein  given. 

Also,  the  weights  of  the  girder  diagonals  will  be  the  same 
as  given  in  that  article. 

We  may  expect  a  heavier  bridge  this  time  than  was  found 
in  the  last  example,  by  reason  of  the  initial  strain  now  assumed 
on  the  top  diagonals,  and  the  smaller  values  of  d  for  the  top 
struts,  the  posts,  and  the  braces,  in  comparison  with  the  values 
used  in  the  two  end  frames  to  resist  wind. 

Computing  weights  for  the  different  values  of  n,  and  col- 
lecting results,  we  have,  — 


/  =  ioo  Foot- Weights  in  Pounds,     IV  and  L  in  Tons,     h  in  Feet,     nL  =  100  Tons. 


n 
5 

Top  chords  j  L°ad  •     • 
<  Initial  st., 

4140.742 

W 
k 

- 

Wh 

82815 

i 

7i 

558 

h" 

- 

Bottom  chords  j  Load  ' 

2444.444 

- 

48889 

- 

- 

<  Wind  . 

— 

— 

— 

— 

122.2222 

Verticals,  total      .     .     . 

- 

9.77778 

- 

- 

568.5926 

Girder  diagonals  .     .     . 

1333-333 

3-33333 

35555 

- 

88.8889 

K  .    . 

- 

- 

- 

49277 

572.6820 

•       2000W  W  — 

7918.519 

+13.11111 

+167259 

+49835 

+  1352.3857 

6 

Top  chords  \  Load   '     ' 

4866.256 

- 

81104 

_ 

- 

(  Initial  St., 

- 

- 

- 

510 

- 

Bottom  chords  {  Load  ' 
(  Wind  . 

2777.778 

_ 

46296 

_ 

115.7408 

Verticals,  total      .     .     . 

- 

14.66667 

- 

- 

544-2387 

Girder  diagonals  .     .     . 

1388.889 

5.00000 

30007 

- 

108.0247 

K  .     . 

- 

- 

- 

45991 

523-30I9 

z<xx>nW  = 

9032.923 

+19.66667 

+157407 

+46501 

+  1291.3061 

7 

Top  chords  j  Load  • 
l  Initial  st. 

5525-323 

; 

78933 

467 

- 

Bottom  chords  j  Load 
(  Wind 

3174.604 

_ 

45351 

- 

"3.3788 

Verticals,  total      .     . 

- 

I9-55556 

- 

- 

489.6447 

Girder  diagonals  .     . 

1360.544 

6.66666 

25915 

- 

126.9841 

K  . 

- 

- 

- 

44005 

485.9160 

•2aaonW  = 

10060.471 

+26.22222 

+150199 

+44472 

+  1215.9236 

370 


MECHANICS  OF   THE    GIRDER. 


1=  zoo  Foot-Weights  in  Pounds,     W  and  L  in  Tons,    h  in  Feet,     nL  =  100  Tons. 


M 
8 

Top  chords  \  Load  •    ' 

6221.067 

_ 

Wh 

77763 

i 

_ 

If 

_ 

h 

'  Initial  St., 

- 

- 

- 

" 

418 

- 

Bottom  chords  j  Load  ' 

3559-027 

- 

44488 

- 

- 

<  Wind  . 

- 

- 

- 

- 

111.2196 

Verticals,  total      .     .     . 

- 

26.07408 

- 

- 

532.7546 

Girder  diagonals  .     .     . 

1388.889 

8.88889 

22787 

- 

I45-8333 

K  .     . 

- 

~ 

- 

42795 

471.2816 

2ooon^  = 

11168.983 

+34.96297 

+145038 

+43213 

+1261.0891 

0 

Top  chords  j  Load  '    ' 

6881.576 

- 

76462 

- 

- 

'  Initial  St., 

- 

- 

- 

394 

- 

Bottom  chords  J  Load  ' 

3978.051 

- 

44201 

- 

- 

f  Wind  . 

- 

- 

- 

- 

110.5014 

Verticals,  total      .     .     . 

- 

32.59260 

- 

- 

513.4888 

Girder  diagonals  . 

1371.742 

ii.  mil 

20322 

- 

164.6090 

K  .     . 

- 

- 

- 

42081 

459.4619 

2ooon  W  = 

12231.369 

+43.70371 

+140985 

-t-42475 

+1248.0611 

* 

TO 

Top  chords  {  Load  '     ' 

7564.819 

- 

75648 

- 

_ 

'  Initial  St., 

- 

- 

- 

364 

- 

Bottom  chords  I  Load  ' 

4388.889 

- 

43889 

- 

- 

<  Wind  . 

- 

- 

- 

- 

109.7222 

Verticals,  total     .    .     . 

- 

40.74075 

- 

- 

578.8889 

Girder  diagonals  .     .     . 

1388.889 

13.88889 

18333 

- 

l83.3333 

K  .    . 

- 

- 

- 

41700 

460.4940 

2OOO«  W  '  = 

13342.597 

+54.62964 

+137870 

+42064 

+1332.4384 

II 

Top  chords  \  Load   '     ' 
t  Initial  St., 

8226.202 

- 

74784 

338 

- 

Bottom  chords  j            ' 
'  Wind  . 

4820.936 

— 

43827 

_ 

109.5668 

Verticals,  total      .     .     . 

- 

48.88888 

- 

- 

572.6966 

Girder  diagonals  .     .     . 

1377.410 

16.66667 

16696 

- 

202.0202 

K  .     . 

- 

- 

- 

41553 

461.4682 

2000«  W  = 

14424.548 

+65.55555 

+135307 

+41891 

+1345.7518 

Top  chords  j  Load  •     ' 

8903.037 

- 

74192 

- 

- 

i  Initial  St., 

- 

- 

- 

315 

- 

Bottom  chords  j  Load  ' 

5246.912 

- 

43724 

- 

-• 

f  Wind  . 

- 

- 

- 

- 

109.3106 

Verticals,  total      .     .     . 

- 

58.66667 

- 

- 

647-5823 

Girder  diagonals  .     .     . 

1388.889 

20.00000 

15325 

- 

220.6790 

K  ..  . 

- 

- 

- 

41574 

471.0348 

2000M  W  = 

15538.838 

+  78.66667 

+133241 

+41889 

+1448.6067 

GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      3/1 

Multiplying  each  of  these  eight  equations  by  — - — ,  we  find 

20000 

the  uniform  panel  weight  of  bridge,  W,  in  terms  of  h ;  thus  : 
8.36295  +  2.49175^  +  0.06761929^ 


«=  5,  W 


n  =  6,  W 


n  =  7,  W  = 


n=  8,  W  = 


—0.3959259  4-  o.5/&  —  0.00065555 
7.87035  +  2.32505^  +  °- 


-0.4516462  +  o.6h  -  0.0009833333^ 

7.50995  4-  2.2236/2  +  0.06079618^ 
—0.5030235  +  0.7^  —  o.ooi3iiiin^2 

7.2519  +  2.16065^  4-  0.06305446^ 
—0.5584492  -f-  o.8/^  —  0.001748  148^" 

7.04925  -|-  2.12375^  4-  0.06240306^ 
—0.61156845  -f  0.9^  —  o.oo2i85i86/^2 

4-  2.1032^  4-  0.06662192^ 


^2      — — •       T  (*\  tfl/        "~"  

—0.667129854-^  —  0.002731482^2" 

n  =  I]:      ^  _     6-76535  +  2.09455^  4-  0.06728759^ 
—0.7212274  -f  i.iA  —  0.003277777/^2' 


ft    =    12 


V 
6.66205  4  2.09445/2  -f-  0.07243034^ 


—0.7769419  4-  1.2/1  —  0.00393333^ 


Differentiating  these  equations,  and  putting  =  o,  we 

ah 

find  results  as  here  tabulated;  h  corresponding  to  the  least 
value  of  n  W. 


372 


MECHANICS  OF   THE   GIRDER. 


Span  /  =  100  Feet,    Uniform  Live  Load  =  nL,  =  100  Tons. 


Number  of  Panels,  n. 

5 

6 

7 

8 

12.60087 

I2.4OQQQ 

I2.^o^6 

1  1  7Q2IO 

Weight  of  bridge,  tons,  n  W  '.  . 
Panel  length  /  *  n  . 

43-5II95 
20  ooooo 

40.91832 
1  6? 

38.99626 
IJ& 

3845946 
i  H 

1  •   k  

7  87Q7O 

8  05800 

8  12640 

8  480^0 

Slope  of  diagonals,  0  .... 
Ratio  of  dead  to  live  load  .  .  . 
Ratio  of  dead  to  total  load  .  . 
Weight  of  bridge  per  lin.  ft.,  Ibs., 
Weight  of  wood,  tons  .... 
Weight  of  iron,  tons  

32°  23'  50" 
0.43510 
0.30320 
SyO.OOOOO 
2O.224OO 
23  288OO 

36°  40'  17" 
0.40920 
0.29040 
818.00000 
18.05700 

22  86lOO 

40°  44'  28" 
0.39000 
0.28050 
770.00000 
16.54700 

22  4.4QOO 

43°  19'  51" 

0.38460 
0.27780 
769.00000 
1  5.44600 

2^01  7OO 

Cost  of  iron,  at  $150  .  .  .  . 
Cost  of  wood,  at  $15  .  .  .  . 
Cost  of  bridge 

$3493  20 
303  36 
•?7n6  c6 

$3429  T5 
270  86 

77OO   OI 

$3367  35 

248    20 
76l  ^    ^H 

#3451  95 
231  69 
^68  i  6A 

Excess  over  least  ...  . 

181  01 

84.  d6 

JW1J    JJ 

o 

68  09 

Cost  per  linear  foot  

37  97 

37  oo 

36  16 

3684 

Number  of  Panels,  «. 

9 

10 

11 

12 

Height  in  feet,  h  . 

II   ^Q2'76 

1  1  067  1  4 

io8-'859 

10  37666 

Weight  of  bridge,  tons,  nW  '  .  . 
Panel  length,  /  —  n  

37-83525 
Ili 

38.08053 

IO  OOOOO 

38.04843 
oJr 

38.60209 

8k 

/  -L.  h  

8.62640 

Q  OI  ^80 

0.27480 

Q  6l7OO 

Slope  of  diagonals,  <j>  .  .  .  . 
Ratio  of  dead  to  live  load  .  .  . 
Ratio  of  dead  to  total  load  .  . 
Weight  of  bridge  per  lin.  ft.,  Ibs., 
Weight  of  wood,  tons  .... 
Weight  of  iron,  tons  

46°  I  2'  51" 
0.37840 
0.27450 
757.00000 
I4.6l^)O 
2~l.22T.OO 

47°54'o" 
0.38080 
0.27580 
762.00000 
13.95900 
24.12200 

49°  59'  8" 
0.38050 
0.27560 
761.00000 

13-435°° 
24  6  1  300 

5i°  13'  57" 

0.38600 
0.27850 
772.00000 
13.00700 

2  e  co  coo 

Cost  of  iron,  at  $150  .  .  .  . 
Cost  of  wood,  at  $15  .... 

$3483  45 
219  18 
•5702  67 

$3618  30 
209  38 
•3827  68 

$369!  95 
201  53 
ogcn  j.8 

^joyjuv-' 

$3839  25 
i95  " 

A.O1A.    T.6 

Excess  over  least 

87  08 

212    17 

/tiS  ST 

Cost  per  linear  foot  

37  03 

38  28 

^11   VJ 

38  93 

4°  34 

GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      373 

Here,  again,  we  find  least  weight,  nW  =  37.83525  tons, 
answering  to  the  odd  number  of  panels,  9,  and  the  height, 
k  =  11.59226  feet ;  while  the  inclination  of  diagonals  to  hori- 
zon, <f>,  is  about  i^  degrees  above  45  degrees. 

The  least  cost,  at  the  rates  here  assumed,  corresponds  to 
7  panels ;  it  being  understood  that  we  have  once  or  twice 
employed  the  approximation  involved  in  (471). 

142.  Again,  by  the  method  of  article  139,  take  the  same 
example,  except  that  the  uniform  live  load  is  now  2  tons  to  the 
linear  foot,  instead  of  i  ton,  as  in  article  141. 

ist,  The  floor,  as  before,  weighs 

F  =  —  X  17.5  X  ioo  x  50  =  18229  pounds. 


12 


2d,  By  (432), 

Thickness  of  a  joist,  b  =  (9  X  10  X  2  x  ioo(l8229  +  4OOooo)  I* 
(  n2  x  17.5  X  7000  ) 

fc2£2"i  inches. 
»t 

And,  from  (433), 


ioo  x   17.^   X   qo/g.O72i?3\        2268^54 

Weight  of  joists,  /  =  -     I447X52    L(9-4^)  =  -^rf  P°unds- 

3d,  Depth  of  I  floor  beams,  from  (412),  as  in  article  140, 

i 


//  +  4i822o\i  . 
=  0.4331885^    -^-   -i)    inches. 

By  (434), 

Weight  of  I-beams, 

c  //  +  418229        18. 

P  =  I5.46o68(«  -  I)  X  ^  X  18.5^   -i-         X 

I  J    +     4l8229\| 

=  i.o3i824(«  -  i)f  -  -  ^—    -  \    pounds. 


3/4  MECHANICS  OF  THE   GIRDER. 

4th,  Take  top  horizontal  diagonals,  each  i^  inches  in  diam- 
eter. Cross-section  =  0.99402  square  inches;  weight  =  3-359 
pounds  per  foot.  Then 

Weight  of  2n  top  horizontal  diagonals 


=  2n  x  3-359V/i82  -f  —  -  =  6.718^324^  +  10000  pounds. 


Weight  of  bottom 
horizontal  diag- 
onals 


_  ^  _  2X5X4X18X6  yy{nZ  (H  even)> 

18  X  24  sin2  </>x         '  (  (n2—  i)  (n  odd), 


as  in  article  140,  for  both  top  and  bottom. 


hwl          i  ,   loooo 

2n'     sin2  iS2n2' 


5th,  Strain  on  each  top  horizontal  strut  from  %£  =  6  tons 
per  square  inch  on  two  top  diagonals  =  2  X  6  X  0.99402 
=   11.928245^^  tons;    allowed   inch   strain   on   strut 

4.680056  , 
—  =  1.170014  tons.     Therefore 


Cross-section  of  a 

top    strut  ^to          ^.92824   . 
resist  initial 
strain  on  diag- 
onals 


. 

sm0,  =   10.1040^ 
1.170014       ri  y^yo  ,    i oooo 

n 


From  (461), 

Gross-section  of  a  top  strut 
to  resist  distorting  force 
of  wind 


2ac  =  "^;"~  =  o.3of^  square  inch. 


GIRDERS    WITH  EQUAL   AND  PARALLEL   CHORDS.      3/5 


From  (465), 


Weight  of  n  top  hori-  J  =  6n.697n*J ? pounds 

zontal  struts  D  *J  V/    \  n2  +  ^0.8642  F 


«2+ 30.8642 


pounds, 


by  reason  of  (471). 


6th,    Weight   to   be   added   to    floor-beams,   due   to   wind, 
=  2  X  P'  in  article  140,  changing  </. 

7th,  Weight  to  be  added  to  all  post^ro  resist  distortion  by 

9000 
wind  =  — —h  pounds,  as  before. 

8th,  Weight  of  all  braces  =  174.667-  pounds,  as  before. 
Computing  for  8  values  of  «,  we  find,  — 

Weights  of  Components  of  K,  in  Pounds.     /  =  100  Feet,    nL  =  200  Tons. 


n. 

5 

6 

7 

8 

Floor    

18229.0000 

18229.0000 

18229.0000 

18229.0000 

Joists 

'^''884  OOOO 

264''''  oooo 

2  1  960  OOOO 

18708  oooo 

(  I  floor  beams      .     .     . 

8303.0000 

9102.0000 

9790.0000 

10398.0000 

(  Do  wind                  . 

4.1.4.4.10/1 

cc.42Q7^ 

64.5618^ 

76.6677/2 

Horizontal  top  struts  < 

2046.0000 

370.2857^ 

2693.ObOO 
308.5714/5 

3354.0000 
264.4898/5 

4019.0000 
231.4286/5 

Horizontal  diagonals  < 

904.0000 
120.6662^ 

989.0000 
125.3704/5 

loSl.OOOO 
125.7336^ 

1178.0000 

133-4035^ 

Braces  

34-9333^ 

2C).IIII/l 

24.9524^ 

21.8333/5 

Residual              .     . 

1  2OO  OOOO 

I2OO.OOOO 

1  200.0000 

1  2OO.OOOO 

f\ 

569.3262^ 

518.4826/5 

479.7376^ 

463.333^ 

\ 

+63566 

+  58635 

+  556l4 

+53732 

376 


MECHANICS  OF   THE    GIRDER. 


Weights  of  Components  of  K ',  in  Pounds.     /  =  100  Feet,    nL  =  200  Tons. 


n. 

9 

10 

11 

12 

Floor 

18220  oooo 

18229  oooo 

18229  oooo 

18229  oooo 

Joists    
(  I  floor  beams      .     .     . 
'  Do.  wind  

16243.0000 
10944.0000 
86.6164% 

14313.0000 
11443.0000 
08.0806^ 

12767.0000 
11903.0000 
109.6167/2 

11501.0000 
12331.0000 

122.  "UVlh 

Horizontal  top  struts  \ 

Horizontal  diagonals  < 
Braces       .... 

4684.0000 
205.7  143/fc 
1279.0000 
138.1040/2 

IQ  A.O74./1 

5347.0000 
185.1429/5 
1383.0000 
147.2220^ 

17.4667^ 

6006.0000 

168.3117^ 

1490.0000 

154.0325^ 

15  8788/2 

6661.0000 

154.2857^ 

1  599.0000 

163-9358/2 

i4.ci;trc;/^ 

Residual    

t;t2OO.OOOO 

1  2OO.OOOO 

I2OO.OOOO 

1  2OO.OOOO 

*{ 

449.842  1  /& 
+  52579 

448.8  2  1  2/$ 
+  5I9I5 

447-S397^ 

+5*595 

455.0907^ 

+  5r52i 

9th,  Taking    Q  =  16.7442   tons,   as   in   article    140,  L  — 

2l       200 ,  , 

—  = tons,  we  now  have 

n          n 


Weight  of  top  chords  due  to  verti-  \  _      5  X  4  X  ioo2     /  ^  ,   2OO\ 

2  X  18  X  i6.y442/z\  n  ) 


cal  pressures,  in  pounds 


—  2 


(n  even), 


X 


4.  3^2  _  2n  —  3 


»  odd). 


Strain  throughout  each  top  chord  due  to  initial  strain  of 
-2?4  X  0.99402  =  5.96412  tons,  along  each  diagonal  between 
top  chords,  is 

5. 964 1 2  cos  $!  tons. 


16.7442 
Allowed  inch  pressure  on  top  chords  =  — - —  =  4.18605  tons. 


GIRDERS   WITH  EQUAL   AND   PARALLEL    CHORDS.      ^77 


Additional   cross-section   of  iron  1        2  x   5.06412 
for  both  top  chords  due  to  in-  j-  =       4.18605      cos 
itial  strain  on  top  diagonals      J 


Additional  weight  for 
top  chords  due  in- 
itial strain  on  top 
diagonals,  pounds 


_  12  x  IPO  X  5  X  284.952  _ 


ioo2 


square  inches. 


94984 

n2  +  i  oooo 


loth,  From  (425), 

ght  of  bottom  cho 
vertical  forces,  pounds 


)  _ 
j 


Weight  of  bottom  chords  due  )  _  5  X  4   X    loo2/^  ,   2Qo\ 

2  x  18  X  24/5  \  n  J 


22n  —  24 


—    3^2    -f    22H    —    21 


Oeven), 


(^odd). 


From  (436),  ^  being  zero,  multiplying  by  2, 

Weight  of  bottom  chords  due  )        5  x  4  X  ioo3  x  0.0225,6 
wind,  in  pounds 


X 


18  x  24  x  18 
-  3«2  •+•  2<2n  —  24 

«3 


(«even), 


Oodd). 


nth,  From  (426),  Q,  being  8. 181818  tons, 
Weight  of  verticals,  in  pounds, 

=    3x5x4   Whl?  |     5  X  4  X  200^   I  in2  +  $n-  io\ 
18x8.1818  2Xi8x8.i8i8\  n  ) 

(n  even), 


3x5x4 
18x8.1818 


5x4X200^ 


2Xi8x8 
(n  odd). 


00^   / 
.i8i8\ 


378 


MECHANICS  OF   THE   GIRDER. 


Weight  of  verticals  due  wind  =  9000— 

n2 


if  d  =  8  inches. 

I  I2 

.I2th,   From  (428),  ~ being  equal  to  I  -\ -, 

cin^  /A  "jj2 /;2 


Weight  of  girder  )  _4X5  X  200 X4/^/^2  — 1\  ,     3x5X4/2 
diagonals         )         i8x24sin2</>  \    n    )     i8X24sin2</> 

(n  even), 

g^A       3x5x4* 


(»odd). 
We  therefore  have,  — 

Weights  in  Pounds,     W  in  Tons,    h  in  Feet,    nL  =  200  Tons. 


n 

5 

Top  chords  j  Load  '     ' 

4140.742 

W 

J 

_ 

Wh 

165630 

T 

_ 

h* 

_ 

f  Initial  st., 

- 

- 

_ 

m 

706 

_ 

Bottom  chords  \  Load  ' 

2444-444 

- 

97777 

- 

<  Wind  . 

- 

- 

- 

- 

122.2222 

o 

Girder  diagonals  .     .     . 

1333-333 

3-33333 

71111 

- 

177-7778 

K  .    . 

- 

- 

- 

63566 

569.3262 

2ooonW  — 

7918.519 

+13.11111 

+334518 

+64272 

+1646.5114 

6 

Top  chords    Load   '     ' 

4866.256 

_ 

162208 

_ 

_ 

Initial  st., 

- 

- 

- 

645 

- 

Bottom  chords  {  Load  ' 

2777.777 

- 

92592 

- 

- 

t  Wind  . 

- 

- 

- 

- 

115-7408 

Verticals      

- 

14.66667 

- 

- 

838.4774 

Girder  diagonals  .     .     . 

1388.889 

5.00000 

60014 

- 

216.0494 

K  .     . 

- 

- 

- 

58635 

518.4826 

2000H  W  = 

9032.922 

+19.66667 

+314814 

+59280 

+1688.7502 

7 

Top  chords  {  Load  '     • 

5525-323 

_ 

157866 

- 

- 

I  Initial  st., 

- 

- 

- 

59° 

- 

Bottom  chords  i  Load  ' 

3174.604 

- 

90702 

- 

- 

f  Wind  . 

- 

- 

- 

- 

113.3788 

Verticals 

19.55556 

795.6160 

Girder  diagonals  .     .     . 

1360.544 

6.66666 

51830 

_ 

253.9682 

K  .    . 

- 

- 

- 

55614 

479-7376 

2000M  W  = 

10060.471 

+26.22222 

+300398 

+56204 

+1642.7006 

GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS,      379 


Weights  in  Pounds,     W  in  Tons,    h  in  Feet,    nL  =  200  Tons. 


;/ 

8 

Top  chords  j  Load  '    ' 
'  Initial  St., 

Bottom  chords  \  Load  ' 
(  Wind  . 

Verticals 

6221.067 
3559-027 

IV 

k 

26.07408 

Wh 

155526 
88976 

i 
~k 

542 

if 

111.2196 
924  8842 

Girder  diagonals  .     .     . 
K  .    . 

1388.889 

8.88889 

45574 

5*732 

291.6666 
463-333I 

2OOOW  W  = 

11168.983 

+34.96297 

+290076 

+54274 

+1791.1035 

9 

Top  chords  j  Load   '     ' 
f  Initial  St., 

Bottom  chords  j  Load  • 
f  Wind  . 

Verticals 

6881.576 
3978.051 

32.59260 

152924 
88402 

499 

110.5014 
915.8665 

Girder  diagonals  .     .     . 

K  .     . 

1371.742 

ii.  mil 

40644 

52579 

329.2180 
449.8421 

2000M  W  = 

12231.369 

+43.70371 

+281970 

+53078 

+1805.4280 

10 

Top  chords  {  Load   '     ' 
I  Initial  St., 

Bottom  chords  1  Load  ' 
<  Wind  . 

Verticals      

7564.819 
4388.889 

40.74075 

151296 
87778 

461 

109.7222 
1067.7777 

Girder  diagonals  .     .     . 
K  .    . 

1388.889 

13.88889 

36667 

5I9I5 

366.6667 
448.8212 

2OOO«  IV  = 

I3342-597 

+54.62964 

+275741 

+52376 

+1992.9878 

ii 

Top  chords  \  Load   '     ' 
'  Initial  st., 

Bottom  chords  j  Load  ' 
'  Wind  . 

Verticals 

8226.202 
4820.936 

48  88888 

149568 
87654 

428 

109.5668 
1071  0130 

Girder  diagonals  .     .     . 
K  .    . 

1377.410 

16.66667 

33392 

51595 

404.0404 
447-8397 

2000M  W  = 

14424.548 

+65.55555 

+270614 

+52023 

+2032.4599 

12 

Top  chords  jL°*d  .     . 
'  Initial  st., 

Bottom  chords  j  Load  • 
'  Wind  . 

Verticals      

8903.037 
5246.912 

58.66667 

148384 
87448 

399 

109.3106 
1232.6646 

Girder  diagonals  .     .     . 
K  .     . 

1388.889 

20.00000 

30650 

51521 

441.3580 
455.0907 

2000«H^  = 

15538.838 

+78.66667 

+266482 

+51920 

+2238.4239 

380  MECHANICS  OF   THE    GIRDER. 

Multiplying  each  of  these  8  equations  by  (Ji  -4-  20000),  we 
find  the  uniform  panel  weight,  W,  of  bridge,  in  terms  of  k, 
thus  : 


16.7259  4-  3.2136/2  4-  0.08232557^ 
n  =    5>     W  — 


n  =    6,     W  = 


=     7, 


=  s, 


n= 


—  0.395926  +  0.5/2  —  o.ooo65555/j2 
15.7407  +  2.964/5  +  o. 


—  0.4516461   +  o.6h  —  0.00098333^ 

15.0199  +  2.8102/2  +  0.08213503/22 

—  0.5030235   -f  0.7^  —  0.001311111^2 

14.5038  +  2.7137/2  +  0.08955518^2 
—0.5584491  +  0.8/1  —  0.001748148/^2 

14.0985  +  2-6539^  4-  0.0902714/22 
—0.6115685  -f-  0.9^  —  0.002185185^2 


13.78705  4-  2.6188/2  4-  0.09964939/22 
—  0.66712985  4-  h  —  0.002731482/22 

13.5307  4-  2.60115^  4*  0.10162299/^2 
—  0.7212274  4-   i.i^  —  0.00327777/22 


=  12, 


13.3241  4-  2.596/2  4-  0.11192120/^2 
-0.7769419  4-  1.2^  —  0.00393333/22 


Differentiating,  and  putting  — - -  =  o,  according  to  equation 

ah 

(470),  we  find,  — 


r 

GIRDERS   WITH  EQUAL   AND   PARALLEL    CHORDS.      381 
HEIGHT,  h,  ANSWERING  TO  MINIMUM  VALUE  OF  nW. 

Span  /  =  100  Feet,    Uniform  Live  Load  nL  =  200  Tons. 


Number  of  Panels,  n. 

5 

6 

7 

8 

I  C.47O76 

14  6l  W^ 

14  72O74 

1  7  471  S4 

Weight  of  bridge,  n  W.  .  ,  . 
Panel  length,  feet,  /  -f  n  .  .  . 
Ratio  of  length  to  height  .  .  . 
Slope  of  diagonals,  0  .  .  .  . 
Ratio  of  dead  to  live  load  .  .  . 
Ratio  of  dead  to  total  load  .  . 
Weight  of  bridge  per  lin.  ft.,  Ibs., 
Weight  of  wood,  tons  .... 

59-97030 
2O.OOOOO 
6.48060 

37°  39'  6" 
0.29985 
0.23068 
1199.00000 
25-55650 

34.4  1  780 

57.05609 

16! 
6.84180 

41°  M'  58" 
0.28528 
0.22196 
1141.00000 
22.32550 
74.7  ^060 

54-55331 
14* 
6.98290 

45°  4'  13" 
0.27277 
0.21431 
1091.00000 
20.99450 
•n.qqSSo 

1O-^-J1j^ 
54.38682 
12* 
744520 

47°  3'  27" 
0.27193 
0.21379 
1088.00000 
18.46850 

^H.QlS^O 

Cost  of  iron,  at  $150  .  .  .  . 
Cost  of  wood,  at  $15  .  .  .  . 
Cost  of  bridge 

$5162  07 

383  35 

ccxr   A'* 

$5209  59 
33488 

r  CAA    47 

$5033  82 
301  42 

e^qe   24 

15387  75 
277  03 
c66i  78 

Excess  over  least  .  .  . 

DJ^J  *K 
2IO    l8 

2OQ   27 

jjjj  •* 

o 

•j^n   CA 

Cost  per  linear'  foot  

55  46 

55  45 

53  35 

5665 

Number  of  Panels,  n. 

9 

10 

11 

12 

Height  in  feet  h  .  . 

I  -5  IO6OI 

12  777OI 

12  O4S7'> 

I  I    40^76 

Weight  of  bridge,  n  W  .  .  .  . 
Panel  length,  feet,  /  -±-  n  .  .  . 
Ratio  of  length  to  height  .  .  . 
Slope  of  diagonals,  0  .... 
Ratio  of  dead  to  live  load  .  .  . 
Ratio  of  dead  to  total  load  .  . 
Weight  of  bridge  per  lin.  ft.,  Ibs., 
Weight  of  wood,  tons  .... 
"Weight  of  iron  tons 

53.61297 

ii| 

7.63010 

49°  42'  33" 
0.26806 
0.21140 
1072.00000 
17.23600 

"?6  777OO 

1'i-JjJUI 

54-43510 

IO.OOOOO 

8.10830 

50°  57'  49" 
0.27218 
0.21394 
1089.00000 
16.27100 
•?3  16410 

54-39900 

9rV 
8.30170 

52°  57'  30" 
0.27199 
0.21384 
1088.00000 
1  5.49800 
•?3  90100 

55-64659 
8* 
8.76900 

53°  50'  33" 
0.27823 
0.21767 
1113.00000 
14.86500 
40  78160 

Cost  of  iron,  at  $150  .  .  .  . 
Cost  of  wood,  at  $15  .... 

#5456  55 
258  54 

C7i  c   oQ 

#5724  62 
244  07 
cq68  60 

#5835  15 
232  47 
6067  62 

£6117  24 

222   98 
6*340   22 

Excess  over  least  
Cost  per  linear  foot  

379  85 
57  15 

633  45 
59  69 

732  38 
60  68 

1004  98 
63  40 

382  MECHANICS  OF  THE   GIRDER. 

Here,  for  weight,  the  minimum  minimorum  is  53.61297  tons, 
n  =  9,  cf)  —  49°  42'  3  3";  while  for  cost,  at  the  assumed  prices, 
the  least  is  $5,335.24,  answering  to  n  =  7,  and  ^  =  45°  4'  13". 

Comparing  these  results  with  the  corresponding  ones  in 
article  141,  we  conclude  :  — 

ist,  For  a  given  span  and  number  of  panels,  if  we  increase 
the  live  load,  we  should  increase  the  height. 

2d,  As  the  live  load  increases,  the  ratio  of  dead  to  both  live 
and  total  loads  diminishes. 

143.  As  another  example,  let  the  span  /=  200  feet;  uniform 
live  load  nL  =  200  tons,  or  i  ton  per  linear  foot  ;  other  data  as 
in  articles  141  and  142.  Compute  for  n  =  8,  9,  10,  n,  12,  13, 


ist,  The  floor  weighs 

F  =  -^  X  17.5  X  200  x  50  =  36458  pounds. 

2d,  By  (432), 

Thickness  of  a  joist,  b  =  \  9  X  I0  X  2  X  20O(36458  +  400000)  i* 
(  n2  X  17.5  X  7000  v*      X  I 

10.151042 

=  —  ^—  —  inches. 
#<M 

And,  from  (433), 

200  x  17.5  X  5o/io.5io423\       705523 
Weight  of  joists,  /  =  -      I44  x  2  --  (      „,.,      )  =  —^-  Pounds. 

3d,  Depth  of  I  floor  beams,  from  (412), 


inches. 


GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      383 

By  (434), 
Weight  of  I-beams, 


-      x      x  I-  x 


=  i.o3i824(«  -  i)  pounds. 

4th,  Top  horizontal  diagonals,  as  in  article  142,  weigh 


h  40000  pounds, 
/  now  being  200. 

Weight  of  bottom  1 

horizontal  diag-  f  =  X  = £ —  JFJ  " 

onals  J  18X24 sin'*,          U«* -  * )  (« odd) . 

Wt  =  **!i,    -^-=i  +42222. 


5th,  Top  horizontal  struts,  as  before,  with  change  of  /  from 
100  to  200. 


+0000 


Cross-section  of  one,  due  initial  strain,  =  S^  =  10.19495 

i 

From  (461), 


h2 

Cross-section  of  one,  due  wind,  =  2ac  =  0.6  il —  square  inches. 

n 


384  MECHANICS  OF   THE   GIRDER. 

From  (465), 
Weight  of  n  top  horizontal  struts 


=  37-02^57^2  +  6ii.697«2V/-  pounds 


=  7405.714-  +  6ii.697«2V/ — —pounds, 

n  \  n2  +  123.4568 

by  reason  of  (471). 

6th,  Weight  to  be  added  to  floor  beams,  due  to  wind,  is 
equal  to 

'  n  (n  even), 


4P'  =  15.416666^ 


£-^-1  (n  odd). 


7th,  Weight  to  be  added  to  posts  to  resist  wind  is  equal  to 

i22  x  5  X  4  X  0.0225  x  2ooA3  72000 

-  ^o  —  -  -  o—  -  -  =  i-8^3  =    —  ^—h  pounds. 
18  X  25  X  8  X  2  n2 

by  reason  of  (471). 

8th,  Braces.     From  (459),  if  b  =  5  feet,  d2  —  4  inches. 

Weight  of  braces 

=  4_XJ  x  5  X  °-0225  X  200^/      ,    ,  _6o^  _  \  =        g       ,a 
18  X   18  X  o.7o7n2         V         3°°°  X  42/ 


716.666  7 
—  —  *> 


by  (471). 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      385 


Computing  for  8  values  of  n,  we  find,  — 

Weights  of  Components  of  K,  in  Pounds.     /  =  200  Feet,    nL  =.  200  Tons. 


n 

S 

9 

10   * 

11 

Floor         

36458.0000 

36458.0000 

36458.0000 

36458.0000 

Joists     

58184.0000 

sou  t;.oooo 

441516.0000 

397  05.  oooo 

(  I  floor  beams     .     .     . 

11294.0000 

11809.0000 

12282.0000 

12721.0000 

(  Do.  from  wind    .     .     . 

i5°-95387z 

170.5809^ 

194.9868^ 

215.9620/5 

Horizontal  top  struts  \ 

2859.0000 
925.7142/5 

3465.0000 
822.8571^ 

4092.0000 
740.5714^ 

4734.0000 
673.2467^ 

Horizontal  diagonals  < 

1656.0000 
527.2219/4 

1729.0000 
504.8319^ 

1808.0000 
502.7783^ 

1891.0000 
495.8917/5 

Braces  

So.sS^ 

79.6296^ 

71.66667* 

6"v  I"?!^ 

Residual 

2OOO  OOOO 

2OOO.OOOO 

2OOO.OOOO 

20OO.OOOO 

K  \ 

1693.4732^ 

I577-8995^ 

I5I0.003I/^ 

I450.25IO>5 

'    \ 

+112451 

+  105976 

+  IOII56 

+97509 

n 

12 

13 

14 

15 

Floor    

36458.0000 

36458.0000 

36458.0000 

36458.0000 

Joists                       « 

35768.0000 

724Q2.OOOO 

2Q727.OOOO 

27^6^.0000 

j  I  floor  beams      .     .     . 

13131.0000 

I35l8.OOOO 

13884.0000 

14231.0000 

1  Do.  from  wind   .     .     . 

240.9957/5 

263.1340^ 

288.8560/; 

312.0545/5 

Horizontal  top  struts  < 

5386.0000 
617.1428/6 

6031.0000 
569.6703^ 

67O8.OOOO 
528.9796^ 

7373.0000 
493.7143/5 

Horizontal  diagonals  < 

1978.0000 
501.4819^ 

2O67.OOOO 
503.6705^ 

2l6l.OOOO 
512.2314/4 

2257.0000 
520.3630^ 

Braces  

597222/5 

55.1282^ 

SI.IQOS/* 

47.7777/5 

Residual    •                   • 

2OOO  OOOO 

2OOO  OOOO 

2OOO  OOOO 

2OOO  OOOO 

K\ 

1419.3426/5 

1391.6030^ 

1381.2575^ 

I373-9095^ 

I 

+94721 

+92566 

+90938 

+89684 

9th,  Taking  Q  =  16.7442  tons,  as  before,  and  L  =  -  =  — 

n        n 

tons,  we  find 


386 


MECHANICS  OF   THE   GIRDER. 


Weight  of  top  chords  due  vertical  )  5  x  4  X  20O2     (  vy  \   2QO\ 

pressures,  in  pounds  (        2  x  18  x  16.7442/1^  n  ) 


-f-  $n  —  2 


(n  even), 


4.  3^2  _  2n  _ 

— 


(0  Odd). 


Strain  throughout  each  top  chord  due  to  initial  strain  of 
-2?4-  X  0.99402  =  5.96412  tons,  along  each  diagonal  between  top 
chords,  is 

5.96412  cos  $!  tons. 


Allowed  pressure  on  top  chords  = 


16.7442 


=  4.18605  tons  per  square  inch. 


.Additional  cross-section  of  iron  for  1 

both  top  chords  due  to  initial  \  =  —  -  777-^  -  cos 

4.18605 
strain  on  top  diagonals 


J 


.Additional  weight  for 
top  chords  due  in- 
itial strain  on  top 
diagonals,  pounds 


'324^  +  40000 
_  12  X  200  X  5  X  569.904  _ 


square  inches. 


4-    2002 


i  oth,  From  (425), 

ight  of  bottom  chor 
vertical  forces,  pounds 


Weight  of  bottom  chords  due  )  _  5   X  4   X   2OO2/  ^  ,    20Q\ 

)        2  x  18  x  24/1  \  n  ) 


—  3«2  4- 


—  24 


(»  even), 


22H   —    21 


(n  odd). 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      387 

From  (436),  t  being  zero,  multiplying  by  2, 

Weight  of  bottom  chords  due  )  _  5  X  4  X  2OO3  x  0.0225^ 

)  =          ~ 


wind,  in  pounds  18  x  24  x  18 

27z3  —  3«2  4-  227Z  —  24 


(n  even), 


2723    _    3^2  4-    2272   —    21 

X  -  *—  -  (n  odd). 


nth,  From  (426),  Q,  being  8.  18181  8  tons, 
Weight  of  verticals  due  load,  pounds, 


_  3  X  5  X  ^n2   ,        5  X  4  X  zoo          ^n2  +  $n  —  io 
18  X  8.181818        2  X  18  X  8.i8i8i8\  n 

(n  even), 


—  3  X  5  X  $Wh(n2  —  i)   ,        5  x  4  X  200/1 


Hn^  —  $n2  —  jn  +  3\ 
\  n2  ) 


i8  X  8.181818  2X18x8.181818 

(n  odd). 

Weight  of  verticals  due  wind,  pounds,  by  (467), 

144  x  5  X  4  X  0.0225  *  200          72000^ 

18  X  2  x  25  X  8  n2 

approximately,  (471). 

T  72 

1 2th,   From  (428),  where =  i  +-L-, 

sin2  <£  ;/2/;2 

Weight  of  girder  diagonals 

—  4  X  5   X   200  X  4/i/n2  —  i\    ,     3X5X4^  -^  2 
18  X  24  sin2  <f>      \      n      /        18  X  24  sin2  <f> 
(n  even), 

_  4  X  5   X  200  X  4/i/n2  —  i\         3x5x4^ 
18  X  24sin2c/>      \      n      /        18  X  24  sin2  (^ 
Oodd). 


388 


MECHANICS  OF   THE   GIRDER. 


Weights  in  Pounds,     W  in  Tons,    h  in  Feet,    nL  =  200  Tons. 


;; 
8 

Top  chords  \  ^ad        • 
<  Initial  St., 

Bottom  chords  j  Load  ' 
<  Wind  . 

24884 

14236 

W 

k 

Wh 

622100 
355900 

i 

7i 

1542 

h" 

889.757 

Girder  diagonals  .     .     . 
K.    . 

5556 

8.8889 

182292 

112451 

291.667 
1693.473 

2000«fF  = 

44676 

+34-9630 

+1160292 

+"3993 

+4784.155 

9 

Top  chords  I  Load  '     ' 
f  Initial  St., 

Bottom  chords  {  Load  • 
i  \V  md  . 

Verticals 

27526 
15912 

611689 
353600 

1476 

884.012 

Girder  diagonals  .     .     . 
K  .    . 

5487 

II.IIII 

162577 

105976 

329.218 
1577.900 

2000W  W  = 

48925 

+43-7037 

+1127866 

+107452 

+4484.774 

TO 

Top  chords  {  Load  •    ' 
i  Initial  St., 

Bottom  chords  {£<• 
Verticals      

30259 
17556 

40.7408 

605180 
351120 

1412 

877.778 
1697.778 

Girder  diagonals  .     .     . 
K  .    . 

5556 

13.8889 

146667 

101156 

366.667 
1510.003 

zooonW  '  = 

53371 

+54.6297 

+1102967 

+102568 

+4452.226 

II 

Top  chords  j  Load  •    ' 
<  Initial  st., 

Bottom  chords  J  Load  ' 
(  Wind  . 

Verticals      

32905 
19284 

48.8889 

598272 
350618 

1350 

876.534 
1591.675 

Girder  diagonals  .     .     . 
K  .     . 

55io 

16.6666 

133567 

97509 

404.040 
1450.252 

•2<xx>nW  = 

57099 

+65-5555 

+1082457 

+98859 

+4322.501 

12 

Top  chords  \  Load   '     ' 
'  Initial  St., 

Bottom  chords  j  Load  • 
'  Wind  . 

Verticals 

35612 
20988 

58.6667 

593533 
3498oo 

1291 

874.486 
1670.165 

Girder  diagonals  .     .     . 
K  .    . 

5556 

20.0000 

122599 

94721 

441-357 
I4I9-343 

2OOO«  W  = 

62156 

+  78.6667 

+1065932 

+96012 

+4405.351 

GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      389 


Weights  in  Pounds,     W  in  tons,    h  in  Feet,    nL  =  200  Tons. 


13 

Top  chords  j  L°ad  •     • 

38260 

IK 

1 

_ 

Wh 

588615 

i 

_ 

*• 

_ 

'  Initial  st., 

- 

- 

- 

"35 

- 

Bottom  chords  J  Load  ' 

22801 

- 

350785 

- 

- 

I  Wind  . 

- 

- 

- 

- 

874.930 

Verticals      

- 

68.4444 

- 

- 

1614.025 

Girder  diagonals  .     .     . 

5523 

23-3333 

113286 

- 

478.633 

K  .    . 

- 

- 

- 

92566 

1391-603 

2000«^  = 

66584 

+91.7777 

+1052686 

+93801 

+4359.191 

M 

Top  chords  ]             '     ' 

40952 

- 

585028 

- 

- 

<  Initial  st., 

— 

- 

— 

1181 

— 

Bottom  chords  J  Load  ' 

24490 

- 

349857 

- 

- 

<  Wind  . 

- 

- 

- 

- 

874-636 

Verticals      

- 

79.8519 

- 

- 

1729.182 

Girder  diagonals  .     .     . 

5556 

27.2222 

105280 

- 

515-874 

K  .     . 

~ 

- 

- 

90938 

1381-257 

zooonW  = 

70998 

+107.0741 

+1040165 

+92119 

+4500.949 

Top  chords  \  Load   '     ' 

43602 

_ 

581360 

_ 

_ 

'  Initial  st., 

- 

- 

- 

1131 

- 

Bottom  chords  J  Load  ' 

26272 

- 

350293 

- 

- 

»  Wind  . 

- 

- 

- 

- 

875.720 

Verticals      

91.2593 

1699.028 

Girder  diagonals  .     .     . 

5531 

31.1111 

98326 

- 

553-o86 

K  .    . 

- 

- 

- 

89684 

1373-909 

2000^  = 

75405 

+122.3704 

+1029979 

+90815 

+4501.743 

Multiplying  each  of  these  8  equations  by  (h  -4-  20000),  we 
find  the  uniform  panel  weight,  W,  of  bridge,  in  terms  of  ht 
thus: 

58.0146  +  5.69965/2  4-  0.2392078^ 


W  = 


=    9, 


n  =  10,     W 


—  2.2338  4-  o.8h  —  o.ooi748i5/z2 
56-3933  +  5-372^  +  0.2242387^ 


—  2.44625  -f-  0.9^  —  o. 

55.14835    -h   5.1284^    +   O.2226II3/Z2 

—  2.66855  H-  h  —  0.002  73  1485^ 


390 


MECHANICS  OF   THE   GIRDER. 


n  =  ii,  W  = 

n  =  12,  JF  = 

«  =  13,  W  =. 

n  =  14,  W  = 

«  =  15,  W  = 


54.12285    +    4.94295/2    +    O.2l6l25/J2 

—  2.88495  -j-  i.i/j  —  O.OO327777/*2 

53.2966  -J-  4.8OO6/!  -4-  O.22O2675/J2 

—  3.1078  +  j.2/i  —  0.00393333/;2 

52.6343  +  4-69005^  -f-  0.2I79595/;2 

—  3.3292  +  1.3/4  —  0.00458888^ 

52.00825  +  4.60595/2  -f-  0.2250475^ 
-3-5499  +  i -4^  -  0.005353705;^ 

51.49895  -f  4-54Q757*  +  0.2250872^ 

—  3.77025  +  1.5^  —  o.oo6u852/j2 


Differentiating  these  equations  according  to  the  form  (470), 
and  solving  for  h  and  W,  we  find  as  follows  :  — 

HEIGHT,  //,  ANSWERING  TO  MINIMUM  VALUE  OF  nW. 

Span  /  =  200  Feet,    Uniform  Live  Load  nL  =  200  Tons. 


Number  of  Panels,  n. 

8 

9 

10 

11 

10.42424 

IQ.4O^H 

IQ.0^2  «D 

18.89406 

Weight  of  bridge,  n  W.     .     .     . 

•*-v7'AT*-T.      ^- 
163.83300 

*  ;7  *r    OOO 

i55-38700 

*  J  ^O    3 

151.80500 

147.73400 

Panel  length,  feet,  /  -f-  n  .    .    . 

25.00000 

22| 

2o.ocxfoo 

ISA 

Ratio  of  length  to  height  .     .     . 

10.29700 

10.30800 

10.50800 

10.58500 

Slope  of  diagonals,  ^     .     .     .     . 

37°  So'  46" 

4i°7/33// 

43°  34'  48" 

46°  5'  54" 

Ratio  of  dead  to  live  load  .    .    . 

0.81900 

0.77700 

0.75900 

0.73800 

Ratio  of  dead  to  total  load    .    . 

0.45030 

0.43720 

0.43150 

0.42490 

Weight  of  bridge,  Ibs.  to  lin.  ft., 

1638.00000 

1554.00000 

1518.00000 

1477.00000 

Weight  of  wood,  tons  .... 

47.32100 

43.48700 

40.48700 

38.08100 

Weight  of  iron,  tons     .... 

116.51200 

111.90000 

111.31800 

109.65300 

Cost  of  iron,  at  $150     .     .    .    . 

$17476  80 

$16785  oo 

$16697  70 

$16447  95 

Cost  of  wood,  at  $15    .... 

709  82 

652  31 

607  31 

5/1    22 

18186  62 

174-27  -Ji 

iT'Soi;  01 

I7OIQ   17 

Excess  over  least                    .     . 

Il6?   4.S. 

/  HO/     O 
4l8    14 

x/Owj    w 

"SS  84 

t        y       I 

o 

Cost  per  linear  foot  

i  *  *^jj    ^.^ 

90  93 

T***          T" 

87  19 

-i<jjj  t_»ij. 
8653 

85  10 

GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      391 


HEIGHT,  h,  ANSWERING  TO  MINIMUM  VALUE  OF  n  W.  —  Concluded. 

Span  /  =  200  Feet,    Uniform  Live  Load  nL  =  200  Tons. 


Number  of  Panels,  n. 

12 

13 

14 

15 

18.4.4.06'? 

18.26650 

17  8o47O 

1  7  W64  ^ 

Weight  of  bridge,  n  W.     .     .     . 

±  W><TT'.XV\) 
147.06400 

145.26000 

A  /  »tJWi|Y  \j 

146.09100 

1  1  oy^T-j 
145.51100 

Panel  length,  feet,  /  -^-  n   .     .     . 

I6| 

ISA 

I4f 

13* 

Ratio  of  length  to  height  .     .     . 

10.84100 

10.94900 

11.23300 

11.36600 

Slope  of  diagonals,  0     .     .     .     . 

47o  S4'  24" 

49°  53'  42" 

51°  1  5'  29" 

52°  50'  52" 

Ratio  of  dead  to  live  load  .     .     . 

0.73500 

0.72600 

0.73000 

0.72800 

Ratio  of  dead  to  total  load     .     . 

0.42370 

0.42070 

0.42210 

0.42110 

Weight  of  bridge,  Ibs.  to  lin.  ft., 

1471.00000 

1453.00000 

1461.00000 

1455.00000 

Weight  of  wood,  tons   .... 

36.11300 

34.47500 

33.09200 

31.91200 

Weight  of  iron,  tons      .... 

110.95100 

110.78500 

112.99900 

113.59900 

Cost  of  iron,  at  $150     .     .     .     . 

$16642  65 

$16617  75 

$16949  85 

$17039  85 

Cost  of  wood,  at  $15     .     .    ... 

54i  7i 

Si?  13 

49638 

478  68 

17184  ^6 

17  1  74  88 

17446   21 

I7Hl8    $1 

Excess  over  least 

/  i^t    3 

i6c  IQ 

*  /  *  Or 

1  1  c  7i 

*•  1  l¥VJ   *j 

Aty    06 

^/j1"  jj 

4QQ    "?6 

*V3    t*f 

**  j   /  * 

<\~  1    w 

T-yy  j  ^ 

Cost  per  linear  foot  •          ... 

8c  cp 

85  68 

87    21 

87  so 

"3   y~ 

"/    *J 

**/  jy 

Of  the  bridge  weights  in  this  case,  the  minimum  minimorum 
is  145.260  tons,  n  =  13,  c/>  —  49°  53' 42";  while  of  the  costs 
at  the  assumed  prices,  the  least  is  $17,019.17,  corresponding  to 
«  =  11,  £  =  46°  5' 54". 

144.  From  articles  141  and  143,  exemplifying  2  bridges  of 
different  spans  but  under  the  same  live  load  per  linear  foot,  we 
may  deduce,  — 

ist,  That,  as  the  length  increases,  the  bridge  weight  per 
linear  foot  increases  ;  or,  the  ratio  of  dead  to  live  load  increases 
nearly  as  the  length. 

2d,  That  the  dead  load  increases  nearly  as  the  square  of  the 
length. 

3d,  That  an  odd  number  of  panels  is  more  favorable  to 
weight  than  an  even  number. 


392 


MECHANICS  OF   THE   GIRDER. 


4th,  That  the  height  of  each  panel  should  be  a  little  greater 
than  its  length. 

5th,  That  the  ratio  of  length  to  height  of  girder  depends 
upon  the  span,  as  well  as  upon  the  live  load,  seen  by  comparing 
articles  141,  142,  143. 

These  principles  are  to  be  seen  in  this  table. 

COMPARATIVE  VIEW  OF  RESULTS. 


Span, 
Feet, 
I. 

Uniform 
Live 
Load, 
Tons, 

Best 
Number 
of 
Panels, 

Best 
Height  of 
Girder, 
Feet, 

Least 
Weight  of 
Bridge, 
Tons, 

Slope  of 
Diagonals, 
*. 

Ratio  of 
Dead 
to  Live 
Load, 

Bridge 
Weight 
per 
Lin.  Ft., 

Ratio  of 
Length  to 
Height, 

7      .      T. 

Panel 
Length, 
Feet, 
7   •   « 

nL. 

n. 

h. 

nW. 

W  +  L. 

Pounds. 

100 

100 

9 

11.592 

37-335 

46°i2/5i" 

0.378 

757 

8.626 

rt| 

100 

200 

9 

13.106 

53-6I3 

49°  42'  33" 

0.268 

1072 

7.630 

ii* 

200 

200 

13 

18.266 

145.260 

49°  S3'  42" 

0.726 

H53 

10.949 

tsA 

These  examples  may  suffice  to  illustrate  a  mode  of  deter- 
mining economical  proportions  for  girders  of  all  classes. 


SECTION  2. 

The  Pratt  Truss  of  Single  System  under  Varying  Live  Load,  without 
taking  Account  of  Wind  Pressure. 

145.  We  shall  here  resume  the  example  of  article  36,  the 
span  being  100  feet  of  10  panels,  and  the  live  load  2  locomotives 
of  given  weight  and  wheel  base. 
Take  n  —  number  of  panels. 

W  =  unknown  panel  weight  of  bridge. 
h  =  20  feet  =  height  of  girders,  pin  to  pin. 
q  =  14  feet  =  width  of  bridge,  in  clear. 
q^  =  1 6  feet  =  width  of  bridge,  extreme. 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      393 

Single  track,  2  rails,  56  pounds  per  yard  each. 
Ties,  6  X  8  X  84  inches,  spaced  8  inches  in  clear. 
2  track  stringers,  12  X  d  inches  each. 
Ties  and  stringers,  pine,  40  pounds  per  cubic  foot. 
Weight  of  2  rails  =  2  X  100  X  ^  =  3,733  pounds. 

x-     »/      O 

Weight  of  75  ties  =      A      X  75  X  7  X  40  —  7,ooo  pounds. 
144 

Panel  length  of  stringers  =  120  inches. 

Panel  weight  of  rails  =  373  pounds. 

Panel  weight  of  ties  =  700  pounds. 

2  X  weight  on  I  pair  of  drivers  =  42,000  pounds  distributed. 

Maximum  weight  on  2  stringers  =  43,073  pounds  uniformly 

distributed. 
Then,  for  both  stringers, 

b  =  breadth  =  24  inches. 
d  =  height     =15  inches. 

Take  f  =  10  =  factor  of  safety  for  pine. 

B  =  8,000  =  breaking-weight  for  pine. 
From  equation  (52), 

M  =  \wl2  —  \  X  43°73  X  120, 
where  wl  =  43,073  pounds ;  and,  from  (160), 

R  +  f  =  ^Bbd*  =  24  X  8000^ 

__  43073  X   120  X  60 
8  X  24  x  8000    ' 
d  =  14.21  inches. 
Call  d  —  1 5  inches, 

2  x  12  x  15  X  TOO  x  40 
Weight  of  2  stringers  =  =  10000  pounds. 

Suppose  2  wrought-iron  I-beams  suspended  at  each  panel 
joint,  and  assume  the  load  on  these  beams  to  be  concentrated 
at  their  centre. 


394  MECHANICS  OF   THE   GIRDER. 

Greatest  load  on  2  beams, 

From  rails  and  ties,     1073  pounds, 
From  stringers,  1000  pounds, 

From  locomotive,     28612  pounds  (article  36), 

Total,  30685  pounds, 

at  centre  has  the  momental  effect  of  61,370  pounds  uniformly 
distributed  along  the  double  beam. 

Hence,  for  each  single  I-beam,  D  of  (412)  is  equal  to  30,685 
pounds,  and  ql  =  16  feet. 

Take  f  =  6  —  factor  of  safety. 

B  •=•  50,000  pounds. 
From  (412), 

730685  x  16  X  6\j 
d2  =  3.80122^-       5QQQQ 1    =  14-79  inches  =  required  depth. 

From  (413), 

730685  x  16  X  6\f 
Area  of  cross-section  of  i  beam  =  S  =  1.28839!  -    —      QQ —    —  J 

=  19.508  inches. 

Now  the  "heavy  1 5-inch  I-beam"  of  the  Union  Iron  Mills, 
Pittsburgh,  Penn.,  weighs  67  pounds  to  the  foot,  and  its  section 
consequently  =  67  X  -^  =  20.1  inches. 

We  will,  therefore,  use  the  heavy  1 5-inch  beam  of  67  pounds 
to  the  foot. 

Weight  of  9  pairs  15 -inch  I-beams,  67  pounds,  16  feet ; 

2  X  9  X   16  X  67  =  19296  pounds. 
Weight  of  1 1  head  struts,  14  feet,  20  pounds  ; 
ii  x  14  X  20  =  3080  pounds. 

Weight  of  40  horizontal  diagonals,  \\  diameter,  3.359  pounds, 

1 8  feet; 

40  X  18  x  3-359  =  2419  pounds. 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      395 

Weight  of  the  residue, 

10  x  200  =  2000  pounds. 

RECAPITULATION. 

Rails  =     3733  pounds, 

Ties  =     7000  pounds, 

Stringers  =  10000  pounds, 

Beams  =  19296  pounds, 

Head  struts  =     3080  pounds, 
Horizontal  diagonals  =     2419  pounds, 

Residue  =     2000  pounds. 

K  —  47528  pounds. 

For  determining  the  girder  strains  in  this  example,  we  have 
already  found,  article  36,  the  greatest  moments  and  greatest 
differences  of  moment  due  the  given  rolling-load. 

From  (65),  we  have  the  moments  due  W, 

M  =  — r(n  —  r)  =  $Wr(io  —  r), 

.-.    M,  =  45  w  +    478-32,  fft  =  2.25  W  4  23.916  tons; 

M2  =  80 W  4    863.71,  H2  —  4.oolV  4-  43.186  tons; 

M3  =  105  IV  4  1105.88,  HI  =  5.25^  4-  55.294  tons; 

M4  =  i2oW  4  1236.95,  HA  =  6.00 W  +  61.848  tons; 

Ms  =  \2$W  +  1276.38,  HS  =  6.25^  4  63.819  tons. 

ArjY  =  2.2$IV  4  23.916  =  difference  of  horizontal  strains. 
A2Zf  =  1.75^+  19-270 
^H  =  i.2$W  4  14.050 
A4/f  =  0.75  W  4-  10.451 
^H  =  0.25  W  4     7-395 

4.785 

2.579 

0.932 

0.058 


396  MECHANICS  OF   THE   GIRDER. 

Let  <£  =  angle  of  elevation  of  any  diagonal, 

.*.  tan0  =  f§  =  2,  log  tan  <f>  —  0.3010300, 
logsin<£  =  9.9515452, 
log  cos  <£  =  9.6505152. 

In  top  chord,  take  ratio  of  panel  length  to  least  diameter 
=  12]  then,  by  (400), 

18  P 

=  —  of  the  Gordon  formula  =  17.176  tons. 


I  •  I22 


3000 

Take  ratio  of  length  of  vertical  to  its  least  diameter  =  40 ; 
then,  by  (400), 

18 
OI  = =  11.74  tons. 

i  +  ^ 


3000 

Let  T  =  50,000  pounds  =  25  tons  =  limit  of  tension. 

f  —  5  —  factor  of  safety. 
Summing  the  strains  on  the  equal  panel  lengths,  we  have 

Weight  of  top  chords  =  ^(23.75^  +  248.063)  X  10  X  A? 
=  460.93^  +  4815  pounds. 

Calling  the  strain  on  end  panels  of  the  bottom  chord  the 
same  as  the  strain  on  the  adjacent  panel,  we  have  strains  in 
bottom  chord, 

HI   =       2.2$W  +       23.916 

H2  =     2.25^  +     23.916 
Hz  =    4.00  W  +    43.186 

^4=       5-25^+       55-294 

Hs  =    6.00  W  +    61.848 
—  19.75^  +  208.160 


GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      397 

/.     Weight  of  bottom  chords,  ^-(19.75^  +  208.16)  x  10  x  ^ 

=  263.33^  +  2776  pounds. 
Strain  on  a  vertical  =  Z  =  A.//  tan  <£, 
by  the  formulas  for  Class  IX.  ; 

/.     Z,  =    4.50  £F  +  47.832 

Z2  =     3.50^  +  38.540 

Z3  =     2.50  £F  -f-  28.100 

Z4  =     i.$QW  -f-  20.902 

Z5  =    0.50^  +  14-790 

^6  =  _  9-57° 

2Z  =  12.50^  +  159.734 

Weight  of  verticals,  ^(12.5  W  +  159.734)  X  20  X  *§- 

Qt 

=  (I2-5^  +  I59-734)  X  ajfl-  =  709.86^+  9071  pounds, 


where  Z6  is  used  twice  to  provide  resistance  to  lateral  shocks. 
Strain  on  a  diagonal  =  Y  =  kH  H-  cos  <£. 


If  we  call  the  strain  on  each  of  the  first  5  counters  equal  to 
that  on  the  fifth,  from  live  load  alone,  we  have 

yt  =  y2  =  y3  =  y4  =  y5  =  4.785  -  Cos<£, 

^6  =  (  7-395  +  0.25  W)  -5-  cos0, 
Y7  =  (10.541  +  0.75  W)  -r-  cos^, 
y8  =  (14.050  +  1.25  J-F)  -s-  cos</>, 
F9  =  (19.270  -f  iJ$W)  -f-  cos0, 
Ko  =  (23.916  +  £.25^)  -j-  cos<£. 

5F  =  (99.007  +  6.25^)  -^  cos<^>. 

Weight  of  diagonals  =  ^(99.007  +  6.25^)  x  IO  X  IO 
T  3  cos2  $ 

=  416.17?^  +  6601  pounds. 


393 


MECHANICS  OF   THE   GIRDER. 


Weight  of  top  chords 
Weight  of  bottom  chords 
Weight  of  verticals 
Weight  of  diagonals 

Weight  of  girders  =  G 


RECAPITULATION. 

=  460.93^  4-  4815  pounds, 

=  263.33^  +  2776  pounds, 

=  709.86^  -f-  9071  pounds, 

=  416.17^  4-  6601  pounds, 

=  1850.29^  4-  23263  pounds, 
K  =  47528  pounds, 

Weight  of  bridge  =  20>o>o?iW  =  1850.29^  4-  70791  =  G  4  K. 

:.     W  =  3.9004  tons  =  panel  weight. 
Weight  of  bridge  =  nW  =  39.004  tons. 

Substituting  3.9004  for  W  in  the  expressions  for  H,  Y,  and 
Z,  we  have  this  strain  sheet :  — 


MAXIMA  STRAINS,  TONS. 
LOADS  APPLIED  AT  BOTTOM  JOINTS. 


CROSS-SECTIONS  IN  SQUARE  INCHES. 


.27  3.27 


EACH  OF  Two  GIRDERS. 

FIG.  115. 

146.  If  the  dead  and  live  loads  are  applied  at  the  upper 
joints,  instead  of  the  lower  joints,  the  structure  becomes  a  deck 
bridge  ;  and  the  compressions  here  found  for  the  verticals  must 
be  increased  by  the  panel  weight  of  dead  load  plus  the  greatest 
apex  load  from  the  locomotives ;  viz.,  for  each  girder  we  must 
augment  Z  by 

3.9004  4-  14.3063 
=  9.1034  tons. 


GIRDERS    WITH  EQUAL   AND  PARALLEL    CHORDS.      399 

_ j 

147.  As  another  example  of  varying  load  applied  at  the 
lower  joints  of  the  Pratt  Truss,  we  will,  in  accordance  with 
the  practice  of  some  engineers,  assume  a  certain  panel  weight 
of  engine,  of  tender,  and  of  train,  and  determine  the  strains 
thence  resulting,  and  also  the  weight  of  the  bridge. 

Let  us  take  the  example  given  by  Col.  Merrill  for  this  truss 
(see  "  Iron  Truss  Bridges  for  Railroads,"  by  Col.  William  E. 
Merrill,  U.S.A.);  viz.,— 

Span  =  200  feet  =  nc  =  I. 

Length  of  panel     =12.5  feet  =  c. 

Height  of  truss      =  18.75  feet  =  h- 

Number  of  panels  =  16  —  n. 

On  each  of  2  trusses, 

Panel  weight  of  engine  =  17,600  pounds. 

Panel  weight  of  tender  =  16,160  pounds. 

Panel  weight  of  cars       =  13,152  pounds. 

The  engine  is  supposed  to  cover  2  panels,  the  tender  2  pan- 
els, and  the  cars  follow.  We  therefore  have, 

ist  panel  weight  of  moving-load  =  Ws  =  8.800  tons,  engine; 
2d  panel  weight  of  moving-load  =  W^  =  8.800  tons,  engine ; 
3d  panel  weight  of  moving-load  =  W^  =  8.080  tons,  tender ; 
4th  panel  weight  of  moving-load  =  W2  =  8.080  tons,  tender ; 
5th  panel  weight  of  moving-load  =  W^  =  6.576  tons,  cars; 
6th,  etc.,  the  same. 

To  find  the  strains  due  to  this  rolling-load,  we  employ  equa- 
tions (91) ;  and  for  convenience,  after  dividing  by  the  height, 
h  =  18.75  feet,  we  may  let  r,  denote  the  number  of  panel 
weights  of  cars,  and  u  =  the  number  of  panel  weights  of 
engine  and  tender  on  the  girder  at  any  time.  We  shall  then 
have  for  the  different  positions  of  the  load,  by  summing  equa- 
tions (91),  and  putting 

X  =  (»  -  rx  -  i)  W2  +  (n  -  r,  -  2)  Wz 

+  (n  -  r,  -  3)  W^  +  .  .  .  (n  -  rx  -  ») 


400 


MECHANICS  OF   THE   GIRDER. 


=  ri(?<12+   -^  +  (rx  +  i)  W2  +  (r,  +  2)  JF3 


HORIZONTAL  STRAINS  AT  JOINTS  FOR  ROLLING-LOAD. 


h  ' 


, 


+  i) 


/  - 


,  -  2)  JFa  +  (r,  +  2>(»  -  rx  -  2) 
,  -  3)^4  +  ...(rf  +  2)  (»  -  rl-u 


GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      40  1 


,  +  i)  (n  -  rt  -  3)  W2  +  (rt  4-  2)  («  -  r,  -  3) 


+  (r,  +  i)  («  -  r,  -  4)  ^2  +  fr  +  2)  (»  -  r,  -  4)  ^3 
+  (»•.  +  3)(«  -'ft  -4)  ^4  +  ..-(ri+4)(»  ->.^»> 


*  ~  ri  "  ^  ' 


(rx  H-  i)(«  -  rf  -  »     F2 

(r,  4-  2)  («  -  rx  -  «)  ?F3  +  (rf  +  3)  («  -  ^i  -  »)  ^4 

.  .  .  (r,  +  u)(n  -  r,  -   u)Wu 

g  («  -  r,  -  u) 


-  c(n  —  rt  —  u  —  2) 

—  -  :  —  — 

nh 


T7  •    c  V 

J-J-n  —  i   —         ~  -f  • 

nh 


Computing  these  values  of  H  from  the  above  data,  for  every 
position  of  the  train  as  it  advances,  a  panel  at  a  time,  from  left 
to  right,  preceded  by  the  engine  and  tender,  we  find  this  table 
of  horizontal  strains  at  the  joints,  in  tons ;  viz.,  as  also  given 
directly  by  the  set  of  equations  (91),  — 


402 


MECHANICS  OF   THE   GIRDER. 


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GIRDERS   WITH  EQUAL   AND  PARALLEL    CHORDS.      403 

The  blanks  in  this  table  may  be  filled  by  continually  adding 
to  itself  each  number  in  the  right-hand  column.  It  follows, 
therefore,  that  this  right-hand  column  expresses  the  negative 
differences  of  simultaneous  horizontal  strains  at  adjacent  joints 
due  to  rolling-load.  It  is  evident,  that,  in  this  case,  these 
negative  differences  are  numerically  the  greatest  differences  of 
horizontal  strains  at  adjacent  joints,  and  may  therefore  be 
employed  to  find  the  maxima  vertical  and  diagonal  strains  due 
live  load. 

The  table  shows  (as  was  to  be  expected  from  this  load,  but 
contrary  to  the  assumption  made  by  Col.  Merrill)  that  the  hori- 
zontal strains  are  not  maxima  throughout  when  the  foremost 
end  of  the  engine  is  at  the  last  joint,  but  that  the  greater  part 
of  these  strains  reach  their  greatest  values  when  the  foremost 
end  (that  is,  forward  panel  weight)  of  the  engine  is  at  the  four- 
teenth joint. 

Since  we  require  only  the  greatest  horizontal  strains,  we 
need  not  compute  the  whole  table,  but  only  enough  of  the 
higher  values  of  //  to  be  certain  that  we  find  the  highest  at 
each  joint.  In  the  present  example,  it  will  suffice  to  compute 
values  of  H  for  the  positions  of  load  when  r,  =  11,  rx  —  10, 
rl  =.  9,  and  these  for  only  the  last  8  joints,  since  the  first  7 
horizontal  joint  strains  are  smaller  than  the  last  7  by  reason  of 
the  unequal  loading. 

We  have,  then,  the  following  brief  solution  :  — 

i.  For  maxima  differences  of  horizontal  strain  due  live 
load. 


nh  nh\          2 

+  (rt  +  2)  W,  +  (r,  +  3)  ^4  +  (r,  +  4> 


404 


MECHANICS   OF   THE    GIRDER. 


c  =  12.5,    h  =  18.75,    «  =  16,    Wi  =  6.576,    W2  =  fF3  =  8.08,    JF4  =  *F5  =  8.8. 


ri 

JPi 

»*• 

w* 

Wi 

^5 

*«_• 

«ii 

-3 

o 

0 

0 

0 

8.8 

^4X8.8 

0.3! 

—  2 

0 

0 

0 

8.8 

8.8 

(^+A)8.8 

i.i 

—  I 

0 

o 

8.08 

8.8 

8.8 

&x  8.8+^x8.08 

2.17 

O 

0 

8.08 

8.08 

8.8 

8.8 

^X  8.8+^X8.08 

3-571 

I 

6.576 

8.08 

8.08 

8.8 

8.8 

-2S4-  x  8.8+tfV  x  8.08+^4  x  6.576 

5-257i 

2 

6.576 

8.08 

8.08 

8.8 

8.8 

5.257^+27(2x8.08+2x8.8+2x6.576) 

7.212 

3 

6.576 

8.08 

8.08 

8.8 

8.8 

7.212  +^(33.76+3  x  6.576) 

9.440! 

4 

6.576 

8.08 

8.08 

8.8 

8.8 

9.4401+1.401+^x6.576 

n-947 

5 

6.576 

8.08 

8.08 

8.8 

8.8 

1  1.94^+  1.40!+  A-x  6.576 

14.72 

6 

6.576 

8.08 

8.08 

8.8 

8.8 

14.72  +1.40!+  6X0.274 

17.770! 

7 

6.576 

8.08 

8.08 

8.8 

8.8 

17.7701+1.40!+  7X0.274 

21.0954 

8 

6.576 

8.08 

8.08 

8.8 

8.8 

21.095^+1.40!+  8X0.274 

24.694 

9 

6.576 

8.08 

8.08 

8.8 

8.8 

24.694  +1.40!+  9X0.274 

28.5! 

10 

6.576 

8.08 

8.08 

8.8 

8.8 

28.51+1.401+10X0.274 

32-74 

ii 

6.576 

8.08 

8.08 

8.8 

8.8 

32.71^+1.401+11X0.274 

37.134 

2.    For   the   maxima   horizontal   strains   due   live   load,   as 
already  computed  and  tabulated  above. 


GIRDERS    WITH  EQUAL   AND  PARALLEL   CHORDS.      405 


333 

P     P     P 


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406 


MECHANICS  OF  THE   GIRDER. 


It  is  manifest  that  the  labor  of  computing  the  maxima  values 
of  H  would  be  much  lessened  if  we  could  legitimately  assume 
that  the  horizontal  strains  are  greatest  throughout  when  the 
head  of  the  engine  is  at  the  last  joint  or  at  any  particular 
joint. 

To  find  the  strains  due  the  unknown  bridge  weight,  2/2  W, 
we  have  the  panel  weight  of  bridge  on  each  girder  =  W,  and 
find  from  equation  (65),  after  dividing  by  //, 


H  =  igr(,  _  r). 

^T  maximum,  tons. 

(            ISt 

=  o,     N0=     o,|difference> 

0 

5         * 

r  =  i,     #T  =    5    JF, 

5    W  +    37.134 

r  =  2,     .//z  =     9-5^* 

q\W  -\-    68.401-! 

3¥      • 

0 

7*    —~    "2          ^^     ~~    T  t      \V 
O)             3               O 

u 

13   W  +    93.802 

<j      J^/^ 

0 

r  =  4,     &4=  16   W, 

I 

16   W  +  113.816 

1 

2~q"  rr  • 

O 

tr 

r  =  5,    ^5=18^,    ^ 

i8£ZF  +  128.84! 

o 
P- 

r  =  6,     ^6  =  20   ^, 

20   W  •+•  139.05^ 

i    fF. 

r  =  7,     H7=  21    W, 

2\     W   -\-    144.876 

r  =  8,     ^8  =  2i|fF. 

2ii^F  +  146.314! 

GIRDERS    WITH  EQUAL   AND   PARALLEL    CHORDS.      407 


Panel. 

Maximum  Difference  of 
Horizontal 
Strain  =  <\H. 

No. 

Maximum  Vertical 
Strain  =    A//  tan  </> 

Maximum  Diagonal  Strain. 

! 

o          —  5  W 

o  or  5 

2 

0.3!      —  4*  W 

o  or  5 

3 

i.i        —  3!^ 

o  or  5 

4 

2.17    -  3  fr 

o  or  5 

5 

3-571    —  23-ff 

o  or  5 

6 

5.257*  —  i%W 

o  or  5 

7 

7.212    -       fF 

(  7.212    —      Jf)sec0 

8 

9.440!  —    \W 

9 

!(  9-440!-  W} 

(  9.440!  —    *  W)  sec  0 

9 

11.94*    +    fW 

10 

l(ii.94*    +    W 

(11.94*    +    *^)sec<p 

10 

14.72      +       W 

ii 

1(14.72      +      JF) 

(14.72      +       ^T)sec0 

ii 

17-770!+  if  W 

12 

1(17.770!+  i$W) 

(17.770!  +  i!fF)sec0 

12 

21.095*  +  2\W 

13 

§(21.095*  +  23^) 

(21.095*  +  2^^)  seep 

13 

24.694    +  3  JT 

14 

1(24.694   +3  W} 

(24.694    +  3  W)  sec  </> 

14 

28.5!      +  3f^ 

15 

1(28-51     +  3!  W) 

(28.5!      +  3§W)sec0 

15 

32.71*     +  At\W 

16 

1(32.71*.  +4**n 

(32.71*    +4*W)sec0 

16 

37.134    +  5  W 

17 

1(37-134   +  5  ^) 

(37.134    +  5  ^)sec0 

In  the  first  6  panels  we  shall  introduce  counters  of  I  square 
inch  cross-section,  and  therefore  capable  of  resisting  safely  5 
tons  where  theoretically  no  strain  appears.  Also,  we  shall  call 
the  strain  on  the  bottom  chords  in  the  first  panel  equal  to  that 
of  the  second  panel ;  viz., 

$W  +  37.134  tons. 

For  each  panel  length  of  top  chord,  take  ratio  of  length  to 
least  diameter  =15.  Then  the  Gordon  formula  becomes,  equa- 
tion (400), 

P  18 

_  =  (/  = =  16.744  tons  per  square  inch. 

T       I  5 

3000 
=  3.3488  tons  =  allowed  inch  strain  on  top  chords. 


408  MECHANICS  OF   THE   GIRDER. 

For  each  vertical  strut,  take  ratio  of  length  to  least  diam 
eter  =  37.5.     Then 

P  18 

—  =  Q"  =  -  -  =  12.255  tons  per  square  inch. 


3000 

I  2.2  ^  15 

'•  —  —  —  2.451  tons  =  allowed  inch  strain  on  verticals. 

In  tension,  5  tons  allowed.  Factor  of  safety  /  =  5  for  all 
parts  of  girder;  panel  length  of  chords  =  12.5  feet;  length  of 
verticals  =  18.75  feet  ;  length  of  diagonals  =  23  feet  ;  wrought- 
iron,  -f^  pound  per  cubic  inch. 

From  these  data  we  find, 

Weight  of  top  chords 

=  4(124^'    +  872.244) 

X    I2'5  X  I0      =    6171.36  fT+    43411  pounds. 
3.3488  x  3 

Weight  of  bottom  chords 
=  4(107$  W  +  763-064) 

X  "'5  *  I0       =    3588.89  W  +    25435  pounds. 


Weight  of  verticals 

=  4(2iJfF    +  193-3571)  Xf 

18.75  X  I0 
k    2.451  x  3 

Weight  of  diagonals 


=    3238.47^"  +     29585  pounds. 


23  X  io sec  0 

X  - — — — =  221 1.38  JF  +  24539  pounds. 

5   *  3  

Weight  of  2  girders  —  G  =  15210.10^  +  122970  pounds. 

W  is  in  tons. 


GIRDERS   WITH  EQUAL   AffD  PARALLEL    CHORDS.      409 


To  find  the  constant  part  of  the  bridge  weight  =  K. 
2  rails,  200  feet,  56  pounds  per  yard,  =  7,467  pounds. 
Rails  5  feet  between  centres. 
Ties  6X8  inches,  16  inches  between  centres. 

12  x  200  =  jtjo  tieSj  j  feej-  long,  40  pounds  per  cubic  foot. 
16 

Weight  of  ties  —  150  X  X  7  X  40  =  14,000  pounds. 

144 

2  track  stringers,  each  15  X  i8J  inches,  40  pounds  per  cubic 

„    V     T  C    V     T  8  1- 

foot,  -  —  -  ?"  X  200  X  40  =  30,833  pounds. 
144 

Depth  of  stringer  is  thus  found  :  — 

Length  between  bearings  =12.5  feet. 

Uniform   load  =  -^   of   rails    and   ties  =  ^ffi3-  =  1,342 

pounds. 
Concentrated  load  =  panel  weight  of  engine  =  2  X  17,600 

=  35,200  pounds,  which  is  equivalent  to  uniform  load  of 

70,400  pounds. 

.-.     Uniform  load  on  2  stringers  30  inches  wide  =  71742  pounds. 

For  pine,  take  factor  of  safety  =  10,  and  the  ultimate  inch 
resistance  to  cross-breaking  B  =  8,000  pounds  ; 

.-.     Moment  due  external  forces  =  J  X  71742  x  12  X  12.5, 
from  equation  (52). 
Moment  of  resistance  due  internal  forces,  equation  (160), 

8oo°  x  3V, 


6 
which  becomes,  after  introducing  the  factor  of  safety, 

8000  X 

OO 


410  MECHANICS  OF   THE   GIRDER. 

Equating  moments  of  external  and  allowable  internal  forces, 
we  find 

J  x  71742  x  12  x  12.5  = 


/.     Depth  =  d  =  18.338  inches,  called  i8J. 

15  pairs  I-beams,  heavy  15-inch,  67  pounds,  16  feet,  =  32,160 
pounds. 

Depth  of  beam  is  thus  determined  : 

Panel  weight  of  rails  =      467  pounds, 

Panel  weight  of  ties  ==       875  pounds, 

Panel  weight  of  stringers  =     1927  pounds, 

Panel  weight  of  engine  =  35200  pounds. 

Weight  on  each  pair  of  I-beams  =  38469  pounds. 

Now,  this  weight  is  actually  concentrated  at  two  points  5 
feet  apart,  under  the  rails,  each  point  being  5  J  feet  from  the 
nearer  end  of  the  I-beams. 

The  moment  at  the  centre  of  the  double  beam  is  therefore, 
according  to  equation  (43), 

Mc  =  2  x  19234.5  x  T%-  x  sJ  x  12  =  \W  x  16  x  12, 

by  equation  (46),  if  Wr  is  the  weight  at  centre  producing  an 
equivalent  moment  ; 

.-.     W  =  26447.4, 

and  52,895  pounds  is  the  equivalent  load  uniformly  distributed 
for  2  beams.  Therefore  uniform  load  on  I  beam  is  26,447 
pounds.  And,  if  6  is  the  factor  of  safety  for  beams,  equation 
(412)  gives,  B  being  =  50,000, 

726447  X  16  X  6\i 

Depth  of  beam  =  d2  =  3.80122!  -   -  )   =  14.076  inches. 

\         50000          / 


GIRDERS    WITH  EQUAL    AND   PARALLEL    CHORDS.      4!  I 

From  (413), 

726447  X   16  X  6\f 

Area  =  1.288391 r      7: )  =  17.667  square  inches. 

\  ^  oooo  / 

1 1\2 

Area  of  similar  beam  iq  inches  deep  =  17.667  x — 

14.0762 

=  20.062  square  inches. 

Area  of  heavy  15 -inch  67-pound  beam  of  the) 

Union  Iron  Mills,  Pittsburgh,  Penn.,  f  =  2<XI  scluare  mches- 

Hence  we  may  with  safety  use  this  beam. 

17  head  struts,  14  feet,  25  pounds,  =  5,950  pounds. 

64  horizontal  diagonals,  I J-  diameter,  19^  feet,  3.359  pounds, 

=  4,192  pounds. 
Residue,  200  pounds  per  panel,  =  3,200  pounds. 

RECAPITULATION. 

Weight  of  rails  =      7467  pounds, 

Weight  of  ties  =    14000  pounds, 

Weight  of  stringers  =    30833  pounds, 

Weight  of  I-beams  =    32160  pounds, 

Weight  of  head  struts  =      5950  pounds, 

Weight  of  horizontal  diagonals  =      4192  pounds, 
Weight  of  residue  =      3200  pounds. 

K  =    97802  pounds. 

Weight  of  girders  =  122970  +  15210.10^. 

Weight  of  bridge  =  220772  +  15210.1  W  pounds 

=  2nWtons  =  2  x  16  x  2000 
=  64000^. 

.'.     48789.9^  =  220772,     W  —  4.52493  tons. 
Weight  of  bridge  =  32^*=  144.798  tons. 


412 


MECHANICS  OF   THE   GIRDER. 


Substituting  this  value  of    W  in  the   expressions  already 
found  for  greatest  strains,  we  have,  — 

MAXIMA  STRAINS,  IN  TONS,  FOR  EACH  OF  Two  GIRDERS. 


152.626 


LOADS  APPLIED  AT  BOTTOM 
FIG.  1  1  6. 


To  find  the  allowed  cross-sections  in  square  inches,  we  divide 
strains  in  top  chord  by  3.3488,  strains  in  verticals  by  2.451,  and 
in  bottom  chord  and  diagonals  by  5. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          413 


CHAPTER    X. 

CALCULATION  OF  THE  WEIGHT  OF  BRIDGES  HAVING  GIRDERS  OF 
CLASS  I.,  AND  DETERMINATION  OF  THE  NUMBER  OF  PANELS 
AND  THE  HEIGHT  OF  GIRDER,  WHICH  RENDER  THE  BRIDGE 
WEIGHT  LEAST  FOR  A  GIVEN  SPAN  AND  UNIFORM  LIVE 
LOAD.  —  LIMITING  SPAN  FOUND. 

SECTION  i. 

General  Specifications  for  Iron  Bridges,  issued  in  1879  by  the  New 
York,  Lake  Erie,  and  Western  Railroad  Company.  O.  Chanute, 
Chief  Engineer. 

148.   General  Specifications  for  Iron  Bridges. 

NEW  YORK,  LAKE  ERIE,  AND  WESTERN  RAILROAD  COMPANY. 

1879. 

GENERAL  DESCRIPTION. 

1.  All  parts  of  the  superstructure  shall  be  of  wrought-iron,  except 
bed  plates  and  washers,  which  may  be  of  cast-iron. 

2.  The  following  modes   of  construction  shall  preferably  be  em-  Kinds  of 
ployed :  —  bridges. 

Spans  up  to  17  feet  ....  Rolled  beams. 

Spans  17  to  40  feet  ....  Riveted  plate  girders. 

Spans  40  to  75  feet  ....  Riveted  lattice  girders. 

Spans  over  75  feet   ....  Pin-connected  trusses. 

In  calculating  strains,  the  length  of  span  shall  be  understood  to  be 
the  distance  between  centres  of  end  pins  for  trusses,  and  between  centres 
of  bearing-plates  for  all  beams  and  girders. 


414 


MECHANICS   OF   THE    GIRDER. 


Spacing  of 
girders. 


Head  room. 


Floor. 


Loads. 


3.  The  girders  shall  be  spaced  (with  reference  to  the  axis  of   the 
bridge)  as  required  by  local  circumstances,  and  directed  by  the  chief 
engineer  of  the  railroad  company.* 

4.  In  all  through  bridges,  there  shall  be  a  clear  head  room  of  20  feet 
above  the  base  of  the  rails. 

5.  The  wooden  floor  will  consist  of  transverse  floor  timbers  extend- 
ing the  full  width  of  the  bridge,  supporting  the  rails  and  guard  beams. 
Their  scantling  will  vary  with  circumstances.     They  will  be  furnished 
and  put  on  by  the  railroad  company. 

6.  Bridges  shall  be  proportioned  to  carry  the  following  loads :  — 

ist,  The  weight  of  iron  in  the  structure. 

2d,  A  floor  weighing  400  pounds  per  lineal  foot  of  track,  to  con- 
sist of  the  rails,  ties,  and  guard  timbers  only. 
These  two  items  taken  together  shall  constitute  the  "  dead  load." 
3<3;  A  moving-load  for  each  track,  supposed  to  be  moving  in  either 
direction,  and  consisting  of  two  "  consolidation  "  engines 
coupled,  followed  by  a  train  weighing  2,240  pounds  per 
running  foot ;   this  "  live  load  "  being  concentrated  upon 
points  distributed  as  in  the  diagram  on  p.  415. 

The  maximum  strains  due  to  all  positions  of  the  above  "  live  load," 
and  of  the  "  dead  load,"  shall  be  taken  to  proportion  all  the  parts  of  the 
structure. 

7.  To  provide  for  wind  strains  and  vibrations,  the  top  lateral  bracing 
in  deck  bridges,  and  the  bottom  lateral  bracing  in  through  bridges,  shall 
be  proportioned  to  resist  a  lateral  force  of  450  pounds  for  each  foot  of 
the  span ;  300  pounds  of  this  to  be  treated  as  a  moving-load. 

The  bottom  lateral  bracing  in  deck  bridges,  and  the  top  lateral 
bracing  in  through  bridges,  shall  be  proportioned  to  resist  a  lateral  force 
of  1 50  pounds  for  each  foot  of  the  span. 

Temperature.        8.  Variations  in  temperature  to  the  extent  of  150  degrees  shall  be 
provided  for. 

*  Generally,  in  through  bridges,  the  clear  width  between  trusses  shall  be  15  feet  for 
single  track,  and  28  feet  for  double  track.  In  deck  bridges,  and  for  the  floor  system  of  all 
bridges,  the  spacing  between  tlte  centres  of  trusses  and  girders  shall  generally  be  as 
follows :  — 


Stresses. 


Lateral 
stresses. 


DOUBLE 

TRACK. 

DESCRIPTION. 

SINGLE  TRACK. 

2  Trusses. 

3  Trusses. 

Deck  truss  bridges  .  .  . 
Deck  plate  girders  .... 
Floor  stringers  

12  feet  or  over. 
8  feet  or  over. 
8  feet  or  over. 

16  feet  or  over. 
16  feet  or  over. 
10  feet  or  over. 

10  feet  or  over. 
10  feet  or  over. 
8  feet  or  over. 

The  centres  of  beams  and  plate  girders  shall  be  not  less  than  4  feet  (on  either  side) 
from  the  centre  of  the  broad  gauge  track. 

The  standard  distance  between  centres  of  tracks  is  13  feet. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


4'5 


e 


e 

e 
e 
e 


e- 
e- 


e- 


9.  All  parts  shall  be  so  designed  that  the  strains 
coming  upon  them  can  be  accurately  calculated, 

10.  Strain  sheets  and  a  general  plan  showing  the  Plans  and 
dimensions   of  the  parts  and  general   details  must  strain  sheets, 
accompany  each  proposal. 

11.  Upon  the  acceptance  of  a  proposal,  a  full  set 
of  working  drawings  must  be  submitted  for  approval 
by  the  chief  engineer  of  the  railroad  company  before 
the  work  is  commenced. 

12.  Unless  otherwise  specified,  the  form  of  truss  Form  of 
may  be  selected  by  the  builder;  but,  to  secure  uni-  truss, 
formity  in  appearance,  it  is  desired  that  all  "  through  " 
trusses  shall  be  built  with  inclined  end  posts. 

13.  In  comparing  competitive  plans,  the  relative 
cost  of  the  wooden  floors  required  will  be  taken  into 
consideration. 

14.  The  following  clauses  are  all  intended  to  apply 
to  iron  construction.     Parties  proposing  to  substitute 
steel  for  particular  parts  will  be  required  to  furnish 
evidence  of  its  strength,  elasticity,  uniformity  in  pro- 
duction, and  adaptability  to  the  intended  purpose. 

PROPORTION  OF  PARTS. 

I.  All  parts  of  the  structures  shall  be  so  propor-  Tensile 
tioned  that  the  maximum  strains  produced  shall  in  no  strains, 
case  cause  a  greater  tension  than  the  following :  — 


On  lateral  bracing 

On  solid  rolled  beams,  used  as  cross  floor 

beams  and  stringers 

On  bottom  chords  and  main  diagonals  .  . 
On  counter  rods  and  long  verticals  .  .  . 
On  bottom  flange  of  riveted  cross  girders, 

net  section 

On  bottom  flange  of  riveted  longitudinal 

plate  girders  over  20  ft.  long,  net  section, 
On  bottom  flange  of  riveted  longitudinal 

plate  girders  under  20  ft.  long,  net  section, 
On  floor  beam  hangers,  and  other  similar 

members  liable  to  sudden  loading  .     .    . 


FIG.  117. 


Pounds  per 
Sq.  Inch. 


15000 

IOOOO 

IOOOO 

8000 

8000 
8000 
7OOO 
6OOO 


416 


MECHANICS  OF  THE   GIRDER. 


Compressive         2.  Compression  members  shall  be  so  proportioned  that  the  maximum 
strains.          load  shall  in  no  case  cause  a  greater  strain  than  that  determined  by  the 
following  formulae :  — 


Shearing- 
strains. 


8000 


P=l 


/>=! 


P=  i  4 

-P=the 
L  =  the 
^=the 


L2 


for  square  end  compression  members. 

for  compression  members  with  one  pin  and  one  square 
end. 

for  compression  members  with  pin  bearings. 


20000^?2 

allowed  compression  per  square  inch  of  cross-section. 

length  of  compression  member,  in  inches. 

least  radius  of  gyration  of  the  section,  in  inches. 


3.  The  lateral  struts  shall  be  proportioned  by  the  above  formulae  to 
resist  the  resultant  due  to  an  assumed  initial  strain  of  10,000  pounds  per 
square  inch  upon  all  the  rods  attaching  to  them,  produced  by  adjusting 
the  bridge. 

4.  In  beams  and  girders,  compression  shall  be  limited,  as  follows :  — 


Pounds 
per  Square 
Inch. 

In  rolled  beams  used  as  cross  floor  beams  and  stringers     .     . 
In  riveted  plate  girders  used  as  cross  floor  beams,  gross  section, 
In  riveted  longitudinal  plate  girders  over  20  feet  long,  gross 

1  0000 

6000 

6000 

In  riveted  longitudinal  plate  girders  under  20  feet  long,  gross 
section  .*. 

COOO 

5.  Members  subjected  to  alternate  strains  of  tension  and  compres- 
sion shall  be  proportioned  to  resist  each  of  them.    The  strains,  however, 
shall  be  assumed  to  be  increased  by  an  amount  equal  to  eight-tenths  of 
the  least  strain. 

6.  The  rivets  and  bolts  connecting  all  parts  of  the  girders  must  be 
so  spaced  that  the  shearing-strain  per  square  inch  shall  not  exceed  6,000 
pounds,  nor  the  pressure  upon  the  bearing-surface  exceed  12,000  pounds 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          417 

per  square  inch  of  the  projected  semi-intrados  (diameter  x  thickness  of 
piece)  of  the  rivet  or  bolt  hole. 

7.  Pins  shall  be  so  proportioned  that  the  shearing-strain  shall  not  Bending- 
exceed  7,500  pounds  per  square  inch,  nor  the  crushing-strain  upon  the  strains, 
projected  area  of  the  semi-intrados  (diameter  x  thickness  of  piece)  of 

any  member  connected  to  the  pin  be  greater  than  12,000  pounds  per 
square  inch,  nor  the  bending-strain  exceed  15,000  pounds  per  square 
inch,  when  the  centres  of  bearings  of  the  strained  members  are  taken  as 
the  points  of  application  of  the  strains. 

8.  In  case  any  member  is  subjected  to  a  bending-strain  from  local 
loadings  (such  as  distributed  floors  on  deck  bridges),  in  addition  to  the 
strain  produced  by  its  position  as  a  member  of  the  structure,  it  must  be 
proportioned  to  resist  the  combined  strains. 

9.  Plate  girders  shall  be  proportioned  upon  the  supposition  that  the  Plate  girders., 
bending  or  chord  strains  are  resisted  entirely  by  the  upper  and  lower 

flanges,  and  that  the  shearing  or  web  strains  are  resisted  entirely  by  the 
web  plate. 

10.  The  compression  flanges  of  beams  and  girders  shall  be  stayed 
against  transverse  crippling  when  their  length  is  more  than  thirty  times 
their  width. 

11.  The  unsupported  width  of  any  plate  subjected  to  compression 
shall  never  exceed  thirty  times  its  thickness. 

12.  In  members  subject  to  tensile  strains,  full  allowance  shall  be 
made  for  reduction  of  section  by  rivet  holes,  screw  threads,  etc. 

13.  The  iron  in  the  web  plates  shall  not  have  a  shearing-strain 
greater  than  4,000  pounds  per  square  inch,  and  no  web  plate  shall  be 
less  than  {  inch  in  thickness. 

14.  No  wrought-iron  shall  be  used  less  than  -^  inch  thick,  except  in 
places  where  both  sides  are  always  accessible  for  cleaning  and  painting. 

DETAILS  OF  CONSTRUCTION. 

1.  All  the  connections  and  details  of  the  several  parts  of  the  structure 
shall  be  of  such  strength,  that,  upon  testing,  rupture  shall  occur  in  the 
body  of  the  members  rather  than  in  any  of  their  details  or  connections. 

2.  Preference  will  be  had  for  such  details  as  will  be  most  accessible 
for  inspection,  cleaning,  and  painting. 

3.  The  web  of  plate  girders  must  be  spliced  at  all  joints  by  a  plate 
on  each  side  of  the  web.     T-iron  must  not  be  used  for  splices. 

4.  When  the  least  thickness  of  the  web  is  less  than  one-eightieth  of 
the  depth  of  a  girder,  the  web  shall  be  stiffened  at  intervals  not  over 
twice  the  depth  of  the  g'rder. 

5.  The  pitch  of  rivets  in  all  classes  of  work  shall  never  exceed  6 
inches,  nor  sixteen  times  the  thinnest  outside  plate,  nor  be  less  than 
three  diameters  of  the  rivet. 


MECHANICS  OF   THE   GIRDER. 


6.  The  rivets  used  will  generally  be  \  and  £  inch  diameter. 

7.  The  distance  between  the  edge  of  any  piece  and  the  centre  of  a 
rivet  hole  must  never  be  less  than  1-4-  inches,  except  for  bars  less  than  2-^ 
inches  wide ;  when  practicable,  it  shall  be  at  least  two  diameters  of  rivets. 

8.  When  plates  more  than  12  inches  wide  are  used  in  the  flanges  of 
plate  or  lattice  girders,  an  extra  line  of  rivets,  with  a  pitch  of  not  over  9 
inches,  shall  be  driven  along  each  edge,  to  draw  the  plates  together,  and 
prevent  the  entrance  of  water 

9.  In  punching  plate  or  other  iron,  the  diameter  of  the  dye  shall  in 
no  case  exceed  the  diameter  of  the  punch  by  more  than  -^  of  an  inch. 

10.  All  rivet  holes  must  be  so  accurately  punched,  that,  when  the 
several  parts  forming  one  member  are  assembled  together,  a  rivet  ^ 
inch  less  in  diameter  than  the  hole  can  be  entered,  hot,  into  any  hole, 
without  reaming  or  straining  the  iron  by  "  drifts." 

11.  The  rivets,  when  driven,  must  completely  fill  the  holes. 

12.  The  rivet  heads  must  be  hemispherical,  and  of  a  uniform  size  for 
the   same-sized  rivets  throughout  the  work.     They  must  be  full  and 
neatly  made,  and  be  concentric  to  the  rivet  hole. 

13.  Whenever  possible,  all  rivets  must  be  machine-driven. 

14.  The  several  pieces  forming  one  built  member  must  fit  closely 
together,  and,  when  riveted,  shall  be  free  from  twists,  bends,  or  open 
joints. 

15.  All  joints  in  riveted  work,  whether  in  tension  or  compression 
members,  must  be  fully  spliced,  as  no  reliance  will  be  placed  upon 
abutting  joints.     The  ends,  however,  must  be  dressed  straight  and  true, 
so  that  there  shall  be  no  open  joints. 

1 6.  The  heads  of  eye-bars  shall  be  so  proportioned  that  the  bar  will 
break  in  the  body  instead  of  in  the  eye.     The  form  of  the  head  and  the 
mode  of  manufacture  shall  be  subject  to  the  approval  of    the  chief 
engineer  of  the  railroad  company. 

17.  The  bars  must  be  free  from  flaws,  and  of  full  thickness  in  the 
necks.     They  shall  be  perfectly  straight  before  boring.     The  holes  shall 
be  in  the  centre  of  the  head,  and  on  the  centre  line  of  the  bar. 

1 8.  The  bars  must  be  bored  of  exact  lengths,  and  the  pin  hole  -fa  inch 
larger  than  the  diameter  of  the  pin. 

19.  The  lower  chord  shall  be  packed  as  narrow  as  possible. 

20.  The  pins  shall  be  turned  straight  and  smooth,  and  shall  fit  the 
pin  holes  within  -fa  of  an  inch. 

21.  The  diameter  of  the  pin  shall  not  be  less  than  two-thirds  the 
largest  dimension  of  any  tension  member  attached  to  it.     Its  effective 
length  shall  not  be  greater  than  the  breadth  of  the  foot  of  the  post  plus 
four  times  the  diameter  of  the  pin.     The  several  members  attaching  to 
the  pin  shall  be  packed  close  together,  and  all  vacant  spaces  between 
the  chords  and  posts  must  be  filled  with  wrought-iron  filling-rings. 


Lower 
chords  and 
suspension 
bars. 


Pins. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          419 

22.  All  rods  and  hangers  with  screw  ends  shall  be  upset  at  the  ends,  Upset  screw 
so  that  the  diameter  at  the  bottom  of  the  threads  shall  be  ^  mcn  larger  ends, 
than  any  part  of  the  body  of  the  bar. 

23.  All  threads  must  be  of  the  United  States  standard,  except  at  the 
ends  of  the  pins. 

24.  Floor  beam  hangers  shall  be  so  placed  that  they  can  be  readily  Floor  beam 
examined  at  all  times.     When  fitted  with  screw  ends,  they  shall  be  pro-  hangers, 
vided  with  check  nuts. 

25.  When  bent  loops  are  used,  they  must  fit  perfectly  around  the  pin 
throughout  its  semi-circumference. 

26.  Compression  members  shall  be  of  wrought-iron  of  approved  Compression 
forms.  members. 

27.  The  pitch  of  rivets,  for  a  length  of  two  diameters  at  the  ends,  - 
shall  not  be  over  four  times  the  diameter  of  the  rivets. 

28.  The  open  sides  of  all  trough-shaped  sections  shall  be  stayed  by 
diagonal   lattice   work   at   distances   not   exceeding   the   width    of    the 
member.     The  size  of  bars  shall  be  duly  proportioned  to  the  width. 

29.  All  pin  holes  shall  be  re-enforced  by  additional  material,  so  as  not 
to  exceed  the  allowed  pressure  on  the  pins.     These  re-enforcing  plates 
must  contain  enough  rivets  to  transfer  the  proportion  of  pressure  which 
comes  upon  them. 

30.  Pin  holes  shall  be  bored  exactly  perpendicular  to  a  vertical  plane 
passing  through  the  centre  line  of  each  member,  when  placed  in  a  posi- 
tion similar  to  that  it  is  to  occupy  in  the  finished  structure. 

31.  The  ends  of  all  square-ended  members  shall  be  planed  smooth,  Abutting 
and  exactly  square  to  the  centre  line  of  strain.  joints. 

32.  All   members  must  be   free  from  twists  or  bends.     Portions 
exposed  to  view  shall  be  neatly  finished. 

33.  The  sections  of  the  top  chord  shall  be  connected  at  the  abutting  Splicing  of 
ends  by  splices  sufficient  to  hold  them  truly  in  position.  top  chord. 

34.  In  no  case  shall  any  lateral  or  diagonal  rod  have  a  less  area  than  Lateral 
£  of  a  square  inch.  bracing. 

35.  The  attachment  of  the  lateral  system  to  the  chords  shall  be  thor- 
oughly efficient.     If  connected  to  suspended  floor  beams,  the  latter  shall 
be  stayed  against  all  motion. 

36.  All  through  bridges  with  top  lateral  bracing  shall  have  wrought-  Transverse 
iron  portals  of  approved  design  at  each  end  of  the  span,  connected  diagonal 
rigidly  to  the  end  posts.  bracing. 

37.  When  the  height  of  the  trusses  exceeds  25  feet,  overhead  diagonal 
bracing  shall  be  attached  to  each  post  and  to  the  top  lateral  struts. 

38.  Pony  trusses  and  through-plate  or  lattice  girders  shall  be  stayed 
by  knee  braces  or  gusset  plates  attached  to  the  top  chords,  at  the  ends, 
and  at  intermediate  points  not  more  than  10  feet  apart,  and  attached 
below  to  the  cross  floor  beams  or  to  the  transverse  struts. 


420 


MECHANICS  OF   THE    GIRDER. 


In  all  deck  bridges,  diagonal  bracing  shall  be  provided  at  each  panel. 
In  double-track  bridges,  this  bracing  shall  be  proportioned  to  resist  the 
unequal  loading  of  the  trusses.  The  diagonal  bracing  at  the  ends  shall 
be  of  the  same  equivalent  strength  as  the  end  top  lateral  bracing. 

Bed  plates.  39.  All  bed  plates  must  be  of  such  dimensions  that  the  greatest  press- 

ure upon  the  masonry  shall  not  exceed  250  pounds  per  square  inch. 

Friction  .  40.  All  bridges  over  50  feet  span  shall   have  at  one  end  nests  of 

rollers.  turned  friction  rollers  formed  of  wrought-iron,  running  between  planed 

surfaces.  The  rollers  shall  not  be  less  than  2  inches  diameter,  and  shall 
be  so  proportioned  that  the  pressure  per  lineal  inch  of  rollers  shall  not 
exceed  the  product  of  the  square  root  of  the  diameter  of  the  roller,  in 
inches,  multiplied  by  500  pounds  (500^). 

41.  Bridges  less  than  50  feet  span  will  be  secured  at  one  end  to  the 
masonry,  and  the  other  end  shall  be  free  to  move  by  sliding  upon  planed 
surfaces. 

Camber.  42.  All  bridges  will  be  given  a  camber  by  making  the  panel  lengths 

of  the  top  chord  longer  than  those  of  the  bottom  chord  in  the  proportion 
of  \  of  an  inch  to  every  10  feet. 


Tension 
tests. 


QUALITY  OF   IRON. 

1.  All  wrought-iron  must  be  tough,  fibrous,  and  uniform  in  character. 
It  shall  have  a  limit  of  elasticity  of  not  less  than  26,000  pounds  per 
square  inch. 

Finished  bars  must  be  thoroughly  welded  during  the  rolling,  and  free 
from  injurious  seams,  blisters,  buckles,  cinder  spots,  or  imperfect  edges. 

2.  For  all  tension  members,  the  muck  bars  shall  be  rolled  into  flats, 
and  again  cut,  piled,  and  rolled  into  finished  sizes.     They  shall  stand 
the  following  tests  :  — 

3.  Full-sized  pieces  of  flat,  round,  or  square  iron,  not  over  4^  inches 
in  sectional  area,  shall  have  an  ultimate  strength  of  50,000  pounds  per 
square  inch,  and  stretch  12^  per  cent  in  their  whole  length. 

Bars  of  a  larger  sectional  area  than  4^  square  inches,  when  tested  in 
the  usual  way,  will  be  allowed  a  reduction  of  1,000  pounds  per  square 
inch  for  each  additional  square  inch  of  section,  down  to  a  minimum  of 
46,000  pounds  per  square  inch. 

4.  When  tested  in  specimens  of  uniform  sectional  area  of  at  least  £ 
square  inch  for  a  distance  of  10  inches  taken  from  tension  members 
which  have  been  rolled  to  a  section  not  more  than  4^  square  inches,  the 
iron  shall  show  an  ultimate  strength  of  52,000  pounds  per  square  inch, 
and  stretch  18  per  cent  in  a  distance  of  8  inches. 

Specimens  taken  from  bars  of  a  larger  cross-section  than  4^  inches 
will  be  allowed  a  reduction  of  500  pounds  for  each  additional  square 
inch  of  section,  down  to  a  minimum  of  50,000  pounds. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          421 

5.  The  same-sized  specimen  taken  from  angle  and  other  shaped  iron 
shall  have  an  ultimate  strength  of  50,000  pounds  per  square  inch,  and 
elongate  15  per  cent  in  8  inches. 

6.  The  same-sized  specimen  taken  from  plate  iron  shall  have  an  ulti- 
mate strength  of  48,000  pounds,  and  elongate  15  per  cent  in  8  inches. 

7.  All  iron  for  tension  members  must  bend  cold,  for  about  90  degrees,  Bending- 
to  a  curve  whose  diameter  is  not  over  twice  the  thickness  of  the  piece,  tests, 
without  cracking.     At  least  one  sample  in  three  must  bend  180  degrees 

to  this  curve  without  cracking.  When  nicked  on  one  side,  and  bent  by 
a  blow  from  a  sledge,  the  fracture  must  be  nearly  all  fibrous,  showing  but 
few  crystalline  specks. 

8.  Specimens  from  angle,  plate,  and  shaped  iron  must  stand  bending 
cold  through  90  degrees,  and  to  a  curve  whose  diameter  is  not  over  three 
times  its  thickness,  without  cracking. 

When  nicked  or  bent,  its  fracture  must  be  mostly  fibrous. 

9.  Rivets  and  pins  shall  be  made  from  the  best  double-refined  iron. 

10.  The  cast-iron  must  be  of  the  best  quality  of  soft  gray  iron.  Cast-iron. 
n.  All  facilities  for  inspection  of  iron  and  workmanship  shall  be  Tests. 

furnished  by  the  contractor.  He  shall  furnish,  without  charge,  such 
specimens  (prepared)  of  the  several  kinds  of  iron  to  be  used  as  may  be 
required  to  determine  their  character. 

12.  Full-sized  parts  of  the  structure  maybe  tested  at  the  option  of 
the  chief  engineer  of  the  railroad  company ;  but,  if  tested  to  destruction, 
such  material  shall  be  paid  for  at  cost,  less  its  scrap  value,  to  the  con- 
tractor, if  it  proves  satisfactory.  If  it  does  not  stand  the  specified  tests, 
it  will  be  considered  rejected  material,  and  be  solely  at  the  cost  of  the 
contractor. 

WORKMANSHIP. 

1.  All  workmanship  shall  be  first-class  in  every  particular. 

2.  Abutting  joints  in  truss  bridges  shall  be  in  exact  contact  through- 
out. 

3.  Bars  which  are  to  be  placed  side  by  side  in  the  structure  shall  be 
bored  at  the  same  temperature,  and  of  such  equal  length,  that,  upon  being 
piled  on  each  other,  the  pins  shall  pass  through  the  holes  at  both  ends 
without  driving. 

4.  Whenever  necessary  for  the  protection  of  the  thread,  provision 
shall  be  made  for  the  use  of  pilot  nuts,  in  erection. 

PAINTING. 

1.  All  work  shall  be  painted  at  the  shop  with  one  good  coat  of 
selected  iron-ore  paint  and  pure  linseed-oil. 

2.  In  riveted  work,  all  surfaces  coming  in  contact  shall  be  painted 
before  being  riveted  together. 


422  MECHANICS  OF   THE   GIRDER. 

Bed  plates,  the  inside  of  closed  sections,  and  all  parts  of  the  work 
which  will  not  be  accessible  for  painting  after  erection,  shall  have  two 
coats  of  paint. 

3.  Pins,  bored  pin  holes,  and  turned  friction  rollers  shall  be  coated 
with  white  lead  and  tallow  before  being  shipped  from  the  shop. 

4.  After  the  structure  is  erected,  the  ironwork  shall  be  thoroughly 
and  evenly  painted  with  two  additional  coats  of  paint  mixed  with  pure 
linseed-oil,  of  such  color  as  may  be  directed ;  the  tension  members  being, 
however,  generally  of  lighter  color  than  the  compression  members. 

ERECTION. 

1.  The  railroad  company  will  take  down  the  old  bridge  if  any  exists. 
It  will  furnish  the  lower  falseworks,  or  supporting-trestles,  only.     The 
use  of  these  falseworks  by  the  contractor  shall  be  construed  as  his 
approval  of  them. 

2.  The  contractor  shall  furnish  all  other  staging  (the  plan  and  con- 
struction of  which  must  be  approved  by  the  chief  engineer),  and  shall 
erect  and  adjust  all  the  ironwork  complete. 

3.  The  contractor  shall  so  conduct  all  his  operations  as  not  to  impede 
the  running  of  the  trains  or  the  operations  of  the  road. 

4.  The  contractor  shall  assume  all  risks  of  accidents  to  men  or 
material  during  the  manufacture  and  erection  of  the  bridge. 

ADDITIONAL  STIPULATIONS. 


The  above  specifications  are  approved. 

Chief  Engineer  N.  K,  L.  E.t  &  W.  R.R. 

The  above  specifications  are  accepted. 

Contractor. 

These  specifications  will  be  modified  to  suit  the  character  of 
the  bridges  here  considered. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          423 

SECTION  2. 
The  Brunei  Girder  of  Single  System. 

149.  Take  the  form  of  girder  shown  in  Fig.  16,  Class  I., 
article  49,  where  the  end  lengths  of  top  chord  are  shorter  than 
other  segments.  Bridge  to  have  2  equal  parabolic  double-bow 
girders,  top  and  bottom  chords  of  the  same  curvature.  Floor 
carrying  load  attached  by  vertical  struts  and  suspenders  to 
bottom  and  top  panel  points,  and  in  the  plane  of  the  axes  of 
the  two  girders. 

To  compute  the  dimensions,  let  /  =  span  in  feet,  h  =  height 
of  the  two  equal  parabolas  composing  each  girder  at  the  centre 
of  span.  Take  the  axis  of  x  horizontal,  that  of  y  vertical ;  then 
the  equation  to  the  upper  parabola,  origin  at  top,  is 

x2  =  ay. 
But,  with  origin  at  left  end,  the  equation  is 


2hf        x2\  ,       . 

=  y(-  -  7}  (47*) 


If  there  are  n  equal  panels,  then  the  value  of  y  at  the 
panel  point,  where  x  =  — ,  becomes 

y  =  ^r(*  =• r);  .  (473) 

and  the  whole  height  is 

2y  =  ^*  ~  r)*  (474) 

By  (473), 
For  r  =  o,  y  =  o ; 

r  =  2,  y  =  2/i2n  ~  4 ; 
^^2 


424 


MECHANICS   OF   THE   GIRDER, 


Twice  these  values  of  y  will  be  the  heights  between  curves 
of  parabolas  ;  but,  as  we  shall  here  assume  each  chord  to  be 
polygonal,  that  is,  to  be  straight  from  apex  to  apex,  the  actual 
heights  of  girder  at  these  points  are  manifestly 


4- 


—          ( v<n   -        ^ 
-  ~\2n  3/> 

2/1 

n2 

2h ,,  \ 


hr  =  yr 


_i  4- 


v         2h 
0   =   -T£- 


(475) 


VALUES  OF  s  IN  (475). 


r 

«  =  4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

I 

5 

9 

13 

17 

21 

25 

29 

33 

37 

41 

45 

2 

7 

IS 

23 

31 

39 

47 

55 

63 

7i 

79 

87 

3 

17 

29 

41 

53 

65 

77 

89 

101 

"3 

125 

4 

31 

47 

63 

79 

95 

in 

127 

H3 

159 

5 

49 

69 

89 

109 

129 

149 

169 

189 

6 

7i 

95 

119 

M3 

167 

191 

215 

7 

97 

125 

153 

181 

209 

237 

8 

127 

159 

191 

223 

255 

9 

161 

197 

233 

269 

10 

199 

239 

279 

ii 

241 

285 

12 

287 

Also  we  have 


cos2  a, 


«2/ 


cos2«3 


«2/ 


COS2«2 


COS2«4 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


42$ 


and  generally 


cos2  ar  n2/2 

VALUES  OF  s,  IN  (476). 


(476) 


r 

«=:4 

G 

8 

10 

12 

14 

16 

18 

20 

22 

24 

I 

3 

5 

7 

9 

II 

13 

IS 

I? 

19 

21 

23 

2 

0 

2 

4 

6 

8 

10 

12 

M 

16 

18 

20 

3 

0 

2 

4 

6 

8 

10 

I  ^ 

14 

16 

4 

o 

2 

4 

6 

8 

10 

12 

5 

0 

2 

4 

6 

8 

6 

o 

2 

4 

In  the  same  manner,  we  derive 
ih* 


COS 


i_  =  ,  +  &L(n  _  2y,    _L_  =I+g(»_IO)». 


cos2/?r  «/2  2 

VALUES  OF  e2  IN  (477). 


(477) 


r 

«  =  4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

I 

2 

4 

6 

8 

10 

12 

14 

16 

10 

2O 

22 

2 

o 

2 

4 

6 

8 

10 

12 

M 

16 

18 

3 

o 

2 

4 

6 

8 

10 

12 

M 

4 

o 

2 

4 

6 

8 

10 

5 

o 

2 

4 

6 

6 

o 

2 

426 


MECHANICS  OF   THE   GIRDER. 


I 

COS2 


COS20r 

VALUES  OF  c3  IN  (478). 


r 

«  =  4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

I 

7 

13 

19 

25 

31 

37 

43 

49 

55 

61 

67 

2 

3 

17 

31 

45 

59 

73 

87 

101 

"5 

129 

H3 

3 

5 

27 

49 

7i 

93 

"5 

i37 

i59 

181 

203 

4 

7 

37 

67 

97 

127 

i57 

187 

217 

247 

5 

9 

47 

85 

123 

161 

199 

237 

275 

6 

ii 

57 

103 

149 

i95 

241 

287 

7 

U 

67 

121 

i75 

229 

283 

8 

15 

77 

'39 

2OI 

263 

9 

17 

87 

157 

227 

10 

19 

97 

T75 

ii 

21 

107 

12 

23 

a  denoting  slope  of  any  segment  of  top  chord, 

/? 'denoting  slope  of  any  segment  of  bottom  chord, 

0  denoting  slope  of  any  Z  web  member,  as  shown  in  Fig.  16. 

=  length  of  end  segment  of  top  chord. 


=  length  of  any  other  segment  of  top  chord. 
—  length  of  any  segment  of  bottom  chord. 
=  length  of  any  Z  member. 


72  COS 


n  cos  0 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


427 


Take  gt  =  18  feet  =  extreme  width  of  bridge. 
q  =  1 6  feet  =  width  of  floor. 
0,  =  inclination  to  plane  of  girder,  of  any  horizontal 

diagonal. 
Then 


sn 


(479) 


18 


=  length  of  horizontal  diagonal  supposed  to  reach 


from  end  to  end  of  the  transverse  I  floor  beams. 

150.  To  compute  the  Moments  and  Strains  in  Chords 
due  to  the  Total  Panel  Weight,  W  +  Z,  applied  at  Each 
Apex,  Top  and  Bottom.  —  We  have,  from  equation  (65), 
moments  at  the  vertical  sections  through  these  apices, 


W 


l(n  -  r)r  = 


W 


(480) 


VALUES  OF  s4  IN  (480). 


r 

,.t 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

, 

3 

5 

7 

9 

II 

13 

IS 

17 

19 

21 

23 

2 

4 

8 

12 

16 

20 

24 

28 

32 

36 

40 

44 

3 

9 

15 

21 

27 

33 

39 

45 

51 

57 

63 

4 

16 

24 

32 

40 

48 

56 

64 

72 

80 

5 

25 

35 

45 

55 

65  . 

75 

85 

95 

6 

36 

48 

60 

72 

84 

96 

108 

7 

49 

63 

77 

91 

105 

119 

8 

64 

80 

96 

112 

128 

9 

81 

99 

117 

135 

10 

100 

120 

140 

ii 

121 

143 

12 

144 

428 


MECHANICS  OF   THE   GIRDER. 


Therefore,  since  Hr  =  —  ?,  equations  (475)  and  (480)  give 


H  = 


which  is  the  horizontal  component  of  the  greatest  strain  in 
each  chord,  at  a  point  directly  above  or  below  an  opposite  apex. 
Hence,  in  the  present  case  :  — 

STRAINS  IN  TOP  CHORD. 


STRAINS  IN  BOTTOM  CHORD. 


VALUES  OF  -  =  0.5  -\ ,  IN  (481). 


«  =  4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

0.600000 

0-S55555 

0.538462 

0.529412 

0.523810 

0.520000 

0.517241 

0.515152 

0-513514 

0.512195 

0.511111 

0.571429 

0-533333 

0.521739 

0.516129 

0.512821 

0.510638 

0.509091 

°-  507937  0.507042 

0.5063290.505747 

0.529412 

0.51*241 

0.512195 

0.509434 

0.507692 

0.506494 

0.505618  0.504950 

0.504425  0.504000 

0.516129 

0.510638 

0.507937 

0.5063290.505263 

0.504505  0.503937 

o-  503497  0.503145 

0.510204 

0.507246 

0.505618 

0.504587 

0.503876  0.503356 

0.5029590.502646 

0.507042 

0.5052630.504202 

0-503497  0.502994 

o.  502618  0.502326 

0.505155 

0.504000 

0.503268  0.502762 

0.502392  0.502110 

0-503937 

0.503145  0.502618 

o.  502242  o.  501961 

0.5031150.502538 

0.502146  0.501859 

0.502513  0.502092 

o.  501792 

0.502075 

°-5OI754 

0.501742 

CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          429 

Owing  to  the  peculiar  form  (0.5  -f-  —  Y  these  values  are 
easily  written  from  a  table  of  reciprocals. 

151.  Weights  of  these  Wrought-Iron  Chords. 

W  =  unknown  panel  weight  of  bridge. 
£  =  given  panel  weight  of  live  load. 
Q  =  allowed  inch  strain  in  top  chord. 

T  =  allowed  inch  strain  in  bottom  chord,  all  in  tons,  say. 

p 

—  z=  cross-section  of  top  chord,  square  inches. 

TJ 

—  =  cross-section  of  bottom  chord,  square  inches. 

VOLUME  OF  SEGMENTS  OF  TOP  CHORD,  CUBIC  INCHES. 

24/^6 


Qn  cos2  «/     Qn  cos2  «2'      (?#  cos2  «3       (?«  cos2  «4 
VOLUME  OF  SEGMENTS  OF  BOTTOM  CHORD,  CUBIC  INCHES. 


7«  cos2  /?       7>/  cos2  ft      7«  cos2          Tn  cos2 

)  _ 
j 


Weight  of  top  chords,  )  _  jj_       24^   ZT 
in  pounds,  j        18       ;z^  cos2  a 


18        «<2  A  4     e 

^  x 


A 

l*     I 


by  reason  of  (476)  and  (481). 

In  summing  (484),  it  will  be  seen  that  only  one  of  the  two 
extreme  panels  is  to  be  counted. 


43°                              MECHANICS   OF   THE    GIRDER. 

M 

f  ,  ^ 

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CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


431 


We  shall  here  assume  that  the  ratio  of  the  panel  length  of 
top  chord  to  its  least  radius  of  gyration  is  100,  and  that  sizes 
of  iron  can  be  exactly  fitted  to  meet  the  required  strains.  We 
have,  then,  for  the  segments  of  the  top  chord,  by  our  specifica- 
tions for  columns  with  flat  ends, 


IQO2 
40OOO 


=  3.2  tons  per  square  inch  of  section. 


These  values  of  2-4,  2-V,  and  Q,  put  in  (484),  give, 


Weight  of  top  chords, ) 
in  pounds,  f 


n 

W  +  L 

(o.6ioi2O/2  -f-  0.703  1  3/*2) 

4 

h 

(o.8449o8/2  +  1.050677^) 

6 

(1.092744/2  +  1.40235^) 

8 

(I.345284/2  +  1.75268^) 

10 

(i.6ooi92/2  -f-  2.ioi79/£2) 

12 

(i.85649O/2  -f-  2.45oio^2) 

14 

(2.H3692/2  4-  2.79790^) 

16 

(2.37i52i/2  -h  S-MSST^2) 

18 

(2.629797/2  -f  3.49263^) 

20 

(2.8884i3/2  +  3.83976^) 

22 

(3.147290^  +  4.18679^) 

24 

Similarly  we  find,  from  (477)  and  (481), 

Weight  of  bottom  chords,)  _  jj_  x  24/^   H 
in  pounds,  j        18       nT   cos2/? 


Th 


n2     e 


to  be  summed  as  follows  :  — 


432 


MECHANICS  OF   THE   GIRDER. 


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CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


433 


Since  T  =  5  tons,  we  have,  from  (485), 


Weight    of    bottom)  __  W  4-  Z 
chords,  in  pounds,  j  h 


(o.400ooo/2  4-  0.40000/^2) 

(o.54684i/2  -f  0.65844^) 
(0.703802/2  4-  0.89390^2) 
(0.864473/2  4-  1.12207/^2) 
(1.026993/2  +  i.  347  2  2/*2) 
(i.  190592/2  4-  1.57091/^2) 
(1.354881/2  +  1.79385^) 
(1.519648/2  +  2.0I636/;2) 


(1.684746/2 
(1.850103/2 


2.23863^) 
2.46074/^2) 


(2.015653/2  +  2.68264/^2) 


6 

8 

10 

12 
14 

16 
18 

20 
22 
24 


152.  To  find  the  Greatest  Strains  and  the  Weights  in 
the  Web  System.  —  Calling  Z  =  o  in  (481),  and  taking  first 
differences,  we  find  horizontal  component  of  strain  in  any  girder 
diagonal  due  to  dead  load,  ;/  IV,  thus  : 


W  In .  e, 

A-J 

h    4     e 


(486) 


434 


MECHANICS   OF   THE    GIRDER. 


8 


d         00 

M       m 


f  f  f  1 


o      o      o      o 


o      o      o      o 

+    +    +    + 


o       o       o*       o* 

+     +     +     + 


y   ?   R 


d         0         O         O         O         0 


o      o      o      o 


6      d 

+      + 


?  t  l' 


000 

£   cS; 


&     4 


00  O  A 

H  M  M 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          435 

These  values  of  A^  are  the  alternate  first  differences ;  the 

£ 

only  ones  required  in  this  calculation,  since  the  whole  set  of  Z 
diagonals,  Fig.  16,  is  strained  equally  with  the  whole  set  of 
Y  diagonals  under  the  same  uniform  dead  load  and  same  live 
load  supposed  to  pass  either  way.  And,  moreover,  we  shall 
compute  only  weights  of  diagonals  due  to  the  greatest  com- 
pressive  strains  developed  in  them  ;  since,  although  they  are 
alternately  in  compression  and  tension  under  live  load,  the 
allowed  unit  strain  in  tension,  5  tons  per  square  inch,  is  so 
much  greater  than  that  allowed  in  compression,  that,  if  a  diag- 
onal can  resist  the  compression,  it  certainly  can  resist  the  ten- 
sion coming  upon  it. 

We  shall,  however,  augment  the  cross-section  of  all  girder 
diagonals  by  eight-tenths  of  the  section  computed  to  resist  the 
compressive  strain,  though  not  quite  in  accord  with  our  general 
specifications.  Also,  each  girder  diagonal  is  to  be  rigidly  con- 
nected, at  or  near  its  centre,  with  the  floor  system,  so  that 
virtually  the  unsupported  length  of  these  struts  is  but  half  of 
what  it  otherwise  would  be.  In  these  ways  we  guard  against 
lateral  shocks  received  by  the  girder  diagonals. 

It  may  be  noted  here  that   the  differences,  A—,  in  (486), 

would  all  vanish  if  both  chords  coincided  with  the  parabolic  curve 
throughout.  Also,  these  differences,  being  negative,  increase 
the  counter  strains,  and  diminish  the  main  strains,  due  live  load, 
contrary  to  what  results  when  the  chords  are  horizontal. 

For  the  live  load,  ?iL,  using  the  values  of  hr  in  (475),  and 
taking  Mr  from  equation  (64),  we  have 

*Lr(r  +   i)(«  -  r) 

W--*— *"    «*i 


436  MECHANICS  OF   THE   GIRDER. 

(HL)r  =  the  horizontal  component  of  chord  strain  at  the 
foremost  end  of  live  load,  due  the  same. 
Also,  from  (68)  and  (475),  we  have 

— rr  +  i»  —  r  —  i 


it  «   = 


(HL)r  +  i  =  horizontal  component  of  chord  strain  due  live 
load  at  one  interval  before  its  foremost  end,  and  is  simultaneous 
with  (HL)f  in  (487)  ; 

.-.     A/fc  =  f  x  -(*6-*5),  (489) 

/?        4 

which  is  the  horizontal  component  of  greatest  strain  on  diag- 
onals due  live  load  alone,  and  made  positive,  and  added  to  A//V 
also  made  positive,  gives  the  total  horizontal  component  of 
maximum  diagonal  strain,  as  thus  expressed  : 


(490) 


Here  also  we  require  only  the  alternate  values  of  £5  and  s6  ; 
that  is,  those  values  which  correspond  to  the  instants  when  the 
foremost  panel  weight  of  live  load  is  directly  under  the  upper 
end  of  the  Z  member. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          437 


I    -     I 


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^    1  Ox       «|S 


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1  O         VO      O         U>   >•»>• 


5  is  u 


S  s  SS    ^I 


HIU  SF3  SI? 


£ 


5   &l 


«>&  IIS   s 


n-r  ^  & 


438 


MECHANICS  OF   THE   GIRDER. 


VALUES  OF  «5  —  s6  IN  (490). 


«  =  4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

0.62857 

0.57778 

0.555*8 

0.54269 

0.53480 

0.52936 

0.52539 

0.52236 

0.51999 

0.51806 

0.51647 

o-S1?^ 

0.52058 

0.51686 

0.51393 

0.51179 

0.51018 

0.50896 

0.50799 

0.50721 

0.50657 

Q-49475 

0.50803 

0.50827 

0-50739 

0.50651 

0.50578 

0.50516  0.50466 

0.50424 

0.48466 

0.50242 

0.50440 

0.50444 

0.50413 

0.50378 

0.50344 

0.50315 

0.47895 

0.49923 

0.50218 

0.50276 

0.50277 

0.50263 

0.50246 

0.47529 

0.49716 

0.50076 

0.50167 

0.50189 

0.50189 

0.47273 

o.4957o 

0-49975 

0.50089 

0.50127 

0.47084 

0.49463 

0.49901 

0.50033 

0.46939 

0.49381 

0.49844 

0.46825 

0.49317 

0.46731 

0.62857 

1-09543 

I-5705I 

2.05224 

2-53837 

3-02746 

3-51859 

4.01129 

4-50511 

4-99985 

5-49530 

Since  the  girder  diagonals  are  to  have  their  end  bearings  on 
the  chord  pins,  and  are  to  have  their  centres  attached  to  the 
floor  system,  we  shall  take  the  ratio  of  whole  length  of  diagonal 
to  least  radius  of  gyration  =  100,  and  the  allowed  inch  strain 


+ 


TOO2 
2OOOO 


=  -  tons. 


We  have,  then,  after  multiplying  by  1.8,  as  above  explained, 

Q  A /T" 

Cross-section  of  i  diagonal  =     ' ' 


Volume  of  i  diagonal 

Weight  of  i  diagonal 

Weight  of  all  girder  diag- )     _ 
onals,  in  pounds, 


from  (478)  and  (490). 


<2i  cos  0  k 

I2/  J      I 

>C[UcUC    I 

.8A^ 

llCllCb, 

cubic 
HH 

inches. 

pounds. 
iff 

X 

72 

18 

Ts 

'x    lj 

n 

x  I2/ 

x3-^ 

< 

cos2 
1.8 
5 

(J 

3 

( 

n 

:os2<9 

CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          439 


£ 

^ 

llh 

to 

4 

to 

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0 

1 

+ 

1 

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OJ 

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0 

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00    to      M      O    O     O     M 
OJ       to      O     4*-     O     OJ     Ol 

£  s  s  §  8  a  ^ 

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1      + 

+  +  +  +  i  i  i 

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to          ^i 

oj    vO    ui     O    OJ    ^j     M 

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to 

»0             vO 

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O       10     01     VI     0\ 
Oi     Oi      ON    00     M 
CO  Oi      to      OO   OJ 

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O      OJ        H     VO     NO 
0      ON     00  vg     4>- 

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4*.      ON   Ol      CO  VI     4*. 

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M     NO     \O     4"     OJ      ON 

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4? 
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(OVI        MOJ       MVOO1       H 
V)IOHO4>-4>.>-i|0 
vOOl      MOJ      MVOvboi 
MV1V)       K)       ONIO«4>. 

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^s 

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ONOIOJ      OOJOvO      w 

M4-V1^         §(°°2         °° 

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H        M        M        M        H 

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OJ 

OOVi      OOO     H     ONO     MO 

"S 

H4W4*.      OO      OOOl      ONON 
OJ      ON4«-      MOlOJ      O      OOVO 

01 

H    to    to    to    to    to    M 

4k      10      O      ONO      OOOJ      ONOOH 

4h.tHMtOMtOVlOl4k.O 

Olt-«OViOOONOJOV| 

"t 

I 

VI     O     Oi      H      tO      t04t      M     4v     VI 
ONM      Ovjvl      to      OOONCOO 
4»-OJOJ      to    OJ    4>-      ONONIO      M 

£?   0 

H      to     OJ     *     ^     OJ     OJ      M      H 

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L      O      OOO4*4^OOV|O1      M 
5OJ4^VJ      ONO      OOONO      OO    CO 

M 

M        1 

JOJ      MOJ      10      MUH      ONIO      OO4v 

440 


MECHANICS  OF   THE   GIRDER. 


Placing  the  values  of   2(«5  —  £&)>   3£32(«5  — 

e6),    2  A-4,  and 

* 

2«32A-4,  in  (491),  we  find 

e 

Weight  of  girder  diagonals,  in  pounds, 

M 

_  W 
h 

L 

(o.O32i42/2  4~  o.393?5/z2)  4  — 

(o,767S5/>  + 

2.16563^) 

ii' 

4 

(0.020589/2  4-  0.32779^) 

(0.205393/2  + 

5.15096^) 

6 

(0.015004/2  4-  0.26906^) 

(0.220852/2  4- 

9.32845^) 

8 

(o.OII78l/2  4-  O.22622/J2) 

(o.23o877/2  + 

I4.7IO96/?2) 

10 

(o.oo9688/2  4-  0.19425^) 

(0.237972/2  + 

21.  29814^) 

12 

(o.oo8229/2  4-  O.I7048/!2) 

(o.243278/2  4- 

29.08846/^2) 

14 

(o.oo7i48/2  4-  0.15133/0 

(o.2474Oi/2  4- 

38-.o8o42//2) 

16 

(0.006326^  4-  0.13892^) 

(o.250705/2  4- 

48.27465^) 

18 

(o.oo566o/2  4-  o.i23i3/?2) 

(o.2534i2/2  + 

59.66964^) 

20 

(0.005  1  28/2  +  °-II352^2) 

(0.255654^  4- 

72.26527^) 

22 

(o.oo4686/2  4-  0.10456^) 

(o.257592/2  + 

86.06221^) 

24 

Weight  of  girders  without  head  system,  in  pounds, 

n 

_  w 
h 

(I.042262/2  4-  1.49688^)  +  ^ 

(i.I869o5/2  + 

3.26876^) 

4 

(1.41  2338/2  +  2.03690^) 

(1.597142/2  4- 

6.86oo7/z2) 

6 

(i.8ii55o/2  4-  2.56531^) 

(2.oi7398/2  + 

ii.6247o/^2) 

8 

(2.22I543/2  4-  3.10097/z2) 

(2.440634/2  -f 

17.58571^) 

10 

(2.636873/2  4-  3-64326^) 

(2.865157/24- 

24-747I5/z2) 

12 

(3.05531  1/2  4-  4.19149^) 

(3.290360/2  4- 

33.10947/22) 

r4 

(3-475  72  1/2  +  4.743°8^2) 

(3-7i5974^2  + 

42.67217^) 

16 

(3.897495^  +  5.30065^) 

(4.i4i874/2  + 

53-43638^2) 

18 

(4.32O2O3/2  4-  5-85439^2) 

(4-567955/2  + 

654OO9O/22) 

20 

(4.743644^  +  6.  41402/22) 

(4.994  1  7o/2  4- 

78.565  77/22) 

22 

(5.  167629^  4-  6.97399^) 

(5-420535/2  4- 

92.93164^) 

24 

153.  The  floor ;  to  be  made  of  2^-inch  oak  planks  weighing 
52  pounds  per  cubic  foot.     Width  =  q  —  16  feet  in  (432). 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          44! 

.'.     Weight  of  floor  =  F  =  ^|  X  16  X  52/  =  ^/pounds.      (492) 

154.  The  joists  of  oak;  longitudinal,  spaced  2  feet  between 
centres,  and  consequently  9  in  number,  all  of  equal  size. 

By  (431)  and  (432),  we  have 

Depth  of  joists  =  b*  =  (9  X  10  X  2/^  +  20OO^  finches,  (493) 
( 16  x  10000/2  ) 

calling  B  =  10,000  pounds  =  breaking  inch  strain    for  oak, 
f  =  10  =  factor  of  safety,  g  =  2  feet  =  distance  between 
centres  of  joists. 
And,  from  (433), 

Weight  of  9  joists  =  J  —  -  -j  0.001125-^(7^  +  200o/zZ)  \ 

{I  )  f 

0.001125—  (F  +  2Qoo«Z)  \  °  pounds.    (494) 
n2  }    ' 

155.  The  wrought-iron  I  floor  beams,  ;/  —   I   in  number 
(single  or  in  pairs,  according  to  live  load,  as  shown  below),  each 
bearing   I   panel  weight  of   live  load,  of   floor,  and  of   joists, 
besides  its  own  weight,  which  is  provided  for  as  explained  in 
article  124. 

Also,  the  I-beams  must  bear  the  longitudinal  strain  due  to 
wind  pressure,  and  to  initial  strain  on  the  horizontal  diagonals 
inserted  between  them  in  each  panel. 

Taking  the  proportions  of  the  beam's  section  as  in  article 

D 

124,  and  calling—  =  10,000  pounds  =  allowed  inch  strain  for 
wrought-iron  beams,  (412)  gives 

(  (F  4-  J  -f-  2ooo//Z)  1 8 )  ir 
Depth  of  I-beam  =  </2  =  3.801 22  j-  IOOoo»  ~J   '      (495^ 

in  inches. 


442  MECHANICS  OF   THE   GIRDER. 


Cross-section  of  I-beam,  from  (413),  is 


Weight  of  (n  —  i) 
I-beams,  due  to 


S  —    I07  d*  square  inches. 

1200 


=  P=  15.46068  X 


load,  pounds, 

' 


Ci8(^+/+200o;2Z))| 
=  77.3034(^-1)  j-      ~^^  --  {,(496) 


by  reason  of  (414). 

156.  Or,  in  order  to  avoid  the  complicated  expressions  for 
weight  of  joists  in  terms  of  L  and  /,  we  may  proceed  as 
follows  :  — 

By  equation  (408), 

Weight  of  floor  =  F  =  ulqt  pounds, 

.*.     uqt  —  h  200oZ         =  panel  weight  of  floor  and  live  load, 
n 


-(  uqt-^  +  200oZ  )  =  uniform  load  on  each  panel  length  of  joist. 

Putting  this  weight  for  Iw  in  equation  (52),  we  have,  for 
each  panel  length  of  joist, 

Moment  due  floor  and  live  load,  M  —  -  -{uqt — f-  2OooZ  )— 

8  ^\      n  ]  n 

=  \B>bd*,  (497) 

by  (161)  ;  BT  being  the  allowed  inch  strain. 

.   Now  we  will  take  bd2  ='-S\ 

n 


(d  in  inches,  /  in  feet), 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


443 


S  =  -bd2  square  inches. 


(498) 


That  is,  the  cross-section,  S,  of  a  joist  is  taken  equal  to  -  times 
its  breadth,  b>  multiplied  by  the  square  of  its  depth,  d.  Then, 
calling  g  =  2  feet,  q  =  16  feet,  w  =  52  pounds,  t  =  -  5.  feet, 

Z?,  =  1,000  pounds  per  square  inch  for  oak,  equation  (497) 
becomes,  after  reducing, 

/520/  \ 

£  =  o.oo  1125!  —   -  +  2000  L  J  square  inches,  (499) 


(500) 


=  7-3125^ 


25!  —  -  4-  2OooZ  )/  pounds. 

\  O                1 

°'63375   pound 

n 

n     pounu, 

0.1584375  pound, 

4 

0.1056250  pound, 

6 

0.0792187  pound, 

8 

°-o63375°  pound, 

10 

0.0528125  pound, 

12 

0.0452679  pound, 

14 

0.0396094  pound, 

16 

0.0352083  pound, 

18 

0.0316875  pound, 

20 

0.0288068  pound, 

22 

0.0264063  pound, 

24 

which  are  the  weights  of  the  9  joists  for  oak. 

157.  If,  instead  of  wood,  we  use  wrought-iron  I-beams  to 
support  the  floor  and  load,  we  may  assume  the  beam's  cross- 
section  to  have  such  proportions  (see  manufacturers'  tables) 

that 

/ 

ion 


444 


MECHANICS  OF   THE   GIRDER. 


where  /  =  moment  of  inertia  of  section ;  5  =  area  of  section, 
in  square  inches ;  d  =  depth  of  beam,  in  inches ;  /  =  length 
of  bridge,  in  feet. 

Then,  from  equations  (52)  and  (187),  we  have  moments  of 
external  and  internal  forces, 

M  =  i  x  glF  +  20oo«Z\  x  iz/  =  *BJ  =  lsl_s 

%       q\  n  J         n  d  n    ' 

/.     -  =  — ,  to  be  used  with  makers'  tables, 
d       zBi 

S  =  o.oooi5f  -      — h  2OooZJ  square  inches, 

if  g  •=.  3.2  feet,  q  =  16  feet,  Bl  =  10,000  pounds  per  square 
inch. 


\  (502) 


Weight 


ight  of  6  weight-iron  longi-)  =  ^  =  6 
tudinal  I-beams,  in  pounds,   ) 


x 


x  I2/A  (5<>3) 


/x  =  6Z/  + 


-          , 
-^-     pound, 

0.1300000  pound, 
0.0866666  pound, 
0.0650000  pound, 
0.0520000  pound, 
o-°433333  pound, 
0.0371429  pound, 
0.0325000  pound, 
0.0288889  pound, 
0.0260000  pound, 
0.0236364  pound, 
0.0216667  pound. 


4 

6 

8 

10 

12 
H 

16 
18 

20 
22 
24 


158.  For  the  transverse  I-beams  supporting  the  longitudinal 
I-beams,  the  floor,  and  load,  we  have  on  each,  exclusive  of  its 

own  weight, 

F  +    L  +  200o;zZ 
pounds. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          445 
From  (52)  and  (187), 


Take  B^  =  10,000  pounds  per  square  inch. 

ql  =  1  8  feet  =  entire  length  of  beam. 
We  shall  assume  the  cross-section  of  each  transverse  I-beam 
to  be  such  that 

7  =  2$d  (5°5) 

whether  the  beam  be  rolled  or  made  up  of  plate  and  angle  iron. 
Therefore,  from  (504)  and  (505), 


/ 

S  =  0.000675 


520/  6Z/ 

-  -  4-  —  + 


+  2oooZ.     (506) 


Weight  of  (n  —  i)  transverse  I-beams  due  load,  in  pounds, 
=  (n  —  i)  x  -fs  X  12  x  186* 

/520/       0.5  2/2       6LI  \ 

=  0.0405 (»  -  i)l  —  H ^ h  —  +  20ooZ  ) 


(507) 


=  / 


n 

5.2650  +  /2 

0.003949  +  Z/ 

0.18225  4-  Z 

243 

4 

5.8500 

0.002925 

0.20250 

405 

6 

6.1425 

0.002303 

0.21262 

567 

8 

6.3180 

0.001895 

0.21870 

729 

10 

6.435° 

0.001609 

0.22275 

891 

12 

6.5186 

0.001397 

0.22564 

i°53 

14 

6.5812 

0.001234 

0.22781 

1215 

16 

6.6300 

0.001105 

0.22950 

1377 

18 

6.6690 

O.OOIOOO 

0.23085 

1539 

20 

6.7009 

0.000914 

0.23195 

1701 

22 

6.7275 

0.000841 

0.23287 

1863 

24 

446  MECHANICS  OF   THE   GIRDER. 

159.  In  order  that  the  wind  pressure  may  be  a  function  of 
the  height,  /i,  of  the  girders,  as  it  manifestly  ought  to  be,  we 
will  assume,  for  wind  pressure  against  these  highway  bridges, 
50  pounds  per  square  foot  of  actual  vertical  surface  presented 

by  both  girders,  estimated  at  —  X  -  square  feet,  to  the  run- 
ning-foot of  bridge.     Therefore 

Wind  pressure  per  linear  foot      =  2500-  pounds. 
Wind  pressure  per  panel  length  =  Wt  =  2500-  pounds. 


No  account  is  here  taken  of  vertical  wind  force. 

Let  the  strains  due  this  horizontal  force  of  wind  be  provided 
for  along  the  floor  system  which  is  placed  midway  between  the 
top  and  bottom  chords  (that  is,  insert  horizontal  diagonals  in 
each  panel  between  the  transverse  I-beams),  and  increase  the 
cross-section  already  found  for  these  transverse  beams  by  an 
amount  required  by  the  wind  force ;  and  let  the  longitudinal 
strain  due  wind  be  taken  up  by  2  longitudinal  wrought-iron 
bars,  or  channels,  extending  the  entire  length  of  bridge, 
securely  attached  to  the  outside  of  the  outer  longitudinal  floor 
beams,  to  the  top  of  every  transverse  beam,  and  to  each  girder 
diagonal,  as  already  explained. 

These  two  longitudinal  bars  or  channels,  and  the  two  out- 
side floor  beams,  are  to  be  made  continuous  throughout,  and 
capable  of  resisting  either  tension  or  compression.  It  will  be 
noticed,  that  the  floor  and  load  use  only  one-half  the  capacity  of 
these  two  outside  longitudinal  floor  beams  ;  also,  that  all  floor 
beams,  longitudinal  and  transverse,  are,  from  the  manner  of 
their  loading,  unable  to  deflect  horizontally. 

160.  The  Horizontal  Diagonals,  2*1  in  Number. — To 
provide  for  travelling  gusts  of  wind,  we  shall  here  assume  that 


CALCULATION  OF  THE    WEIGHT  OF  BRIDGES.          447 

this  panel  pressure,  Wt  =  2,500-  pounds,  is  a  uniform  live  load. 

Therefore  the  strains  upon  the  diagonals  are  given  in  Fig.  112 
if  we  put  Wl  in  the  place  of  Z,  and  make  W  =  o. 

But,  since  we  must  provide  for  this  wind  pressure  coming 
upon  either  side  of  the  bridge,  it  is  plain  that  all  horizontal 
diagonals  must  be  "mains,"  and  the  two  in  any  panel  equal  in 

size.     We  must,  therefore,  take  -  terms  of  the  following  series 
four  times  : 

Strain  on  horizontal  diagonals  due  wind 

=  — l^—\n(n-i},  («_i)(«-2),  («-2)(»-3),...  -terms},  (508) 
2»sin<p,  (  2 

in  same  denomination  as  Wv 

Take  the  inch  strain  7^  =  15,000  pounds. 

Weight  of  horizontal  diagonals,  pounds, 


4V                 '  l  •     -^j                 vv                             T 

n(n  —  i)  -f-  (n  —  T)(; 

1-2)' 

1               A\ 

•  x    .     ,    x  m  x        T-    •    j. 
sin  9r                 2w7Ism<pI 

...  -  terms 

2 

<•       4) 

J 

(509) 


since  ;«  =  T5B  pounds  per  cubic  inch  for  wrought-iron,  ql  =  18 
feet  =  length  of  transverse  I-beams. 


_.  J  _l_  / I 

sin2  (j),  vfyi/  ' 

by  (479)- 


448 


MECHANICS  OF   THE   GIRDER. 


From  (509),  we  find 
Weight  of  horizontal  diagonals,  pounds, 


=  h 


hi2 


n 

0.0043403 

4 

0.0029531 

6 

0.0022304 

8 

0.0017901 

10 

0.0014944 

12 

0.0012823 

14 

0.0011225 

16 

0.0009985 

18 

0.0008989 

20 

0.0008174 

22 

0.0007494 

24 

that  is,  in  this   Case,  the 


22.500    -f- 

34-444 
46.250 
58.000 
69.722 
81.429 

93-I25 
104.815 
116.500 
128.182 
139.861 

161.  The  Horizontal  Struts 
Quantity  of  Iron  to  be  added  to  the  Transverse  I-Beams 
by  Reason  of  Wind  Pressure.  —  If  we  divide  the  terms  of 
equation  (508)  by  15,000,  we  shall  have  the  cross-sections  of 
the  horizontal. diagonals  in  square  inches.  And,  if  each  of  these 
sections  be  multiplied  by  1 0,000  sin  <£„  the  product  will  be  the 
longitudinal  pressure  brought  upon  the  end  of  any  transverse 
I-beam  by  one  horizontal  diagonal.  Now,  by  our  specifications, 
the  pressure  so  brought  upon  these  horizontal  struts  by  the 
.horizontal  diagonals  attached  to  the  end  of  each  is  the  end 
pressure  to  be  provided  for  in  these  struts  or  I-beams.  We 
therefore  have 
Longitudinal  pressure  upon  end  of  transverse  I-beams 

«-  s)(*-an  etc) 

«-  2)(«  -  3)J'         I 

=  — [(»  -  i)a,  («  ~  2)2,  («  ~  3)2>  etc.].      (510) 
3« 

These  I-beams  being  unable  to  deflect  horizontally,  and 
having  considerable  depth,  we  may  take  for  them,  under  this 
wind  pressure,  the  unit  strain, 


CALCULATION-  OF   THE    WEIGHT  OF  BRIDGES. 


449 


8000 


8000 


pounds  per  square  inch, 


20000  x 


—l          I    4-    0.933I27 


where  -^  is  put  for  the  square  of  the  radius  of  gyration  about 

an  axis  normal  to  web  of  beam. 

Hence  the  areas  of  sections  to  be  added  to  the  I-beams,  by 
reason  of  wind,  are 

S=-^[(«-  i)2,  (»  ~2)2>  (»  -  3)%  etc.].    (511) 


Taking  m  —  T\,  q,  =  18,  JFZ  =  2,500-,  we  find 

Weight  of  iron  to  be  added  to  transverse  I-beams,  on  account  of  wind, 
in  pounds, 

_     1\2 


_^      (n  -  i)24-  («  -  2)24- 

=  4   X   12 

^7         V 

..(--  1}  terms 

/I      ^                              ' 

\z          / 

2mfj,W,,                             -.    ,             -. 
=  -      '  —  L(7«    —  i2«  4-  2)  (;z  even), 

3<2a 

=  M%  +  0.972^7^  -  12  -f  ^X 
\24                  /A                       */ 

n 

=  ^ 

18.188     4-    - 

64.15 

4 

31-597 

176.91 

6 

46,111 

344.09 

8 

60.625 

565-71 

10 

- 

75-173 

841-75 

12 

89-733 

1172.18 

14 

104.297 

I557'i5 

16 

118.866 

1996.49 

18 

133-438 

2490.27 

20 

148.008 

3038.47 

22 

162.587 

3641.11 

24 

(5") 


450  MECHANICS  OF   THE   GIRDER. 

162.  The  Chords  required  in  the  Horizontal  System  to 
resist  Wind  Force.  —  The  strains  generated  in  these  chords 
by  the  panel  pressure  of  wind,  Wlt  are  given  in  Fig.  1.12  if,  for 

N  =  W  +  Ll,  we  put  N  =  ^,  thus : 
211/1  2Jiql 

.Maxima  chord  strains  in  horizontal  system 

W I 

=  ^-C("  ~  *)>  2(H  ~  2)>  3(»  ~  3),  etc.],       (513) 

for  each  chord,  since  the  wind  may  act  on  either  side  of  the 
bridge. 

Now,  since  these  strains  will  be  sometimes  in  tension  and 
sometimes  in  compression,  these  wind  chords  must  be  con- 
structed so  as  to  resist  either  kind  of  strain.  Then,  of  course, 
a  cross-section  sufficient  for  the  greatest  compressive  strain 
will  be  ample  for  the  maximum  tensile  strain. 

And,  because  these  chords  are  to  be  securely  attached  to 
the  girder  diagonals,  to  the  outside  longitudinal  I-beams  whose 
strength  is  only  one-half  taxed  in  supporting  the  floor  and  live 
load,  and  to  the  transverse  I-beams,  we  shall  take 

unsupported  length 

r: ^ ; —   IOO, 

radius  of  gyration 

as  in  case  of  the  top  chords,  article  151  ;  also,  call  the  ends 
fixed.  The  axis  of  gyration  is  normal  to  the  plane  of  girder, 
since  the  floor  prevents  these  struts  from  deflecting  horizon- 
tally. Then 

Q  =  3.2  tons  =  6400  pounds  =  allowed  inch  strain. 

Cross-section  of  wind  chords  for  each  panel 

=  JZ!\n  -  i,  ,(*  -  »),  3(«  -  3),  -..  ?(«  -  *-)}•      (5M) 
2nqlQ\  2\         2/J 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          451 


Weight  of  wind  chords,  in  pounds, 


=  4  X         X     -- 
1  8         n 


0.006028164/2  -f  ^        4 


«-  a),---      terms 


(5*5) 


=  /z/2 


0.0158239 
0.0147353 
0.0141285 
0.0137442 
0.0134800 
0.0132866 
0.0131395 
0.0130238 
0.0129304 
0.0128534 
0.0127889 


n  = 


4 

6 

8 

10 

12 
14 

16 
18 

20 

22 
24 


163.  The  vertical  supports  for  the  transverse  floor  beams 
and  their  load  must  also  resist  the  moment  due  that  part  of  the 
wind  pressure  which  acts  upon  the  chords  and  girder  diagonals. 

ist,  The  total  weight  to  be  upheld  by  each  of  these  vertical 
struts  and  hangers  is  the  (2;/)th  part  of  the  sum  of  the  weights 
of  the  live  load,  the  floor,  the  longitudinal  I-beams,  the  trans- 


verse I-beams  taken 


times,  the  horizontal  diagonals  in 


11  —  i 

the  floor  system,  and  the  wind  chords ;  all  of  which  have  now 
been  found  in  terms  of  /  and  L.  Call  this  weight  gx  pounds  on 
each  strut  -tmd  on  each  hanger,  ;/  being  the  number  of  panels. 
Each  strut,  of  course,  transmits  the  load,  eM,  to  the  panel  point 
below ;  while  each  hanger  or  suspender  transmits  the  load,  ew,  to 
the  alternate  panel  points  above. 


452  MECHANICS  OF   THE    GIRDER. 


For  the  struts,  we  may  take 
8000 


"  =  5333  P°unds 


2OOOO 

as  the  allowed  inch  strain  in  compression. 

.*.     Cross-section  of  a  strut  due)  ?„ 

....  >   =  o  =  -£  square  inches.       (516) 

vertical  forces  )  Q3 

Weight  of  all  struts  due)  =  2  x  J.  x  ^   x  I22j; 
vertical  forces  )  18        Q3 

(    4^ 

=  o.ooo2o8|f  n \/isn  pounds,        (517) 

since,  from  (473),  for  lower  parabola, 

zh  (                                                                           fn         \  ) 

2v  =  — <  (211  —  4)  +  (AH  —  16)  +  (6/2  —  36)  .  .  .  ( i  |  terms  i 

n2\  f 


Similarly,  for  the  suspenders,  take  T^  =.  6,000  pounds  =  the 
allowed  inch  strain  in  tension. 

Cross-section    of    suspender)  =  s        i»  Q  ^^ 

due  vertical  forces  )  Tt 

Weight  of  all  suspenders  due)2Xj[_x£«x   I22 
vertical  forces  )  18        Tt 

=  0.000185  f*  4-  -V^«  pounds,     (519) 
\         nl 

since,  for  the  upper  parabola, 

-i)+  (3«  -  9)  +  (5«  -  25)  •  •  •  f 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          453 

2cl,  We  shall  assume,  that,  of  the  wind  pressure,  125  pounds 
act  upon  each  running-foot  of  the  two  top  chords  and  of  the  two 
bottom  chords,  tending  to  rotate  each  girder  about  its  longitu- 
dinal axis.  We  may  note,  that,  in  general,  these  forces  act- 
ing upon  one  chord  will  be  nearly  balanced  by  the  forces  acting 
upon  the  other  ;  but  in  certain  cases  a  gale  may  strike  one 
chord  and  not  the  other. 

Acting,  then,  in  lines  normal  to  the  plane  of  each  girder,  at 

2/  /• 

each  panel  point  or  apex,  is  the  pressure  of  62.5  X  --  ='125- 

//  // 

pounds,  with  the  lever  arm,  y,  causing  a  moment,  at  the  wide 

end  of  the  strut  or  suspender,  of  125-  y  foot-pounds.     We  take 

n 

no  account  here  of  the  fact  that  each  end  segment  of  the  top 
chord  is  only  about  one-half  the  length  of  any  other. 

Let  these  struts  and  suspenders,  acting  also  as  lateral  braces 
to  the  chords  where  there  is  no  lateral  head  system,  have  a 
breadth  of  effective  base  equal  to  -£$y.  The  broad  end  of  the 
suspender  is  to  be  attached  to  the  top  chord  and  head  lateral 
strut  whenever  it  would  obstruct  unduly  the  roadway  below. 
Otherwise,  and  in  all  cases  where  head  laterals  are  wanting,  the 
suspender  has  its  broad  end  securely  bolted  to  the  transverse 
I-beam  in  the  floor  system. 

Then,  if  S2  =  cross-section  of  the  two  members  or  flanges 
of  each  strut  or  suspender,  we  have,  at  the  broad  end  of  each, 
this  equality  of  moments, 


75  / 
=  7^  n  square  mches>  (520) 

if  Bt  =  1(5,333  +  6,000)  —  5,667  pounds  =  allowed  inch  strain 
in  bending. 


454  MECHANICS   OF   THE    GIRDER. 

It  will  be  noticed  that  the  cross-section,  S2,  is  uniform 
throughout  the  member  if  the  two  flanges  meet  at  one  end,  as 
we  shall  assume  they  do,  and  shall  illustrate  in  specifications. 

We  have,  then, 

Weight  of  verticals   required   to   resist  bending-moment  due  wind,  in 
pounds, 

=  2  X  78  X  1^'n  X    I2^  =  °-93o3922(i  -  £}Ut   (521) 
since  now 


^y  =  — -[(«  -  i)  4-  2(n  -  2)  +  3(»  —  3)  .  .  .  (»  -  i)  terms] 
=  ^  -  ^. 

From  (517),  (519),  and  (521),  we  find 

Weight  of  all  vertical  supports,  in  pounds, 

^(0.000393518*   -  a0°°f2963)  +  0.9803922(1-^/1     (522) 


(o.ooi  i88Z7+2.3765Z+o.oooooo67373+o.ooooi7i672+ 0.990247 +0.0081) 
(o.ooi2o6Z7+3.2i54Z  +  0.0000006327  3+  o.ooooi3o672+  1.003737  +  0.0152) 
(o.ooi2i4Z7+4.o464Z+o.oooooo6o473+o.ooooio5272+  1.010357+0.0244) 
(o.ooi2i8Z7+4.8733Z+o.oooooo58473+o.ooooo88o72+  1.014707  +  0.0358) 
(o.oo  1 22 1 Z7  +  5-698oZ  +  0.0000005707 3  +  0.000007  5^7  2  +  i  .01 7  587  +  0.0494) 
(o.ooi223Z7 +6.521  iZ+o.oooooo55973+  o.ooooo66272+  1.019887  +  0.0650) 
(o.ooi  224Z7+7-3434Z+  0.0000005  507  3+  o.ooooo58972+  1.021727  +  0.0828) 
(o.ooi225Z7+8.i65oZ+o.oooooo54o73+o.ooooo53i72+  1.023427  +  0.1028) 
(o.ooi225Z7+8.986iZ+o.oooooo53773+o.ooooo48372+  1.024877  +  0.1249) 
(o.ooi226Z7+9.8o69Z+  o.oooooo53i73+o.ooooo44372+  1.026267  +  0.1491)  24 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          455 

since,  for  sn  in  (£22),  we  have 

£4  =  o.7So37Z7  +  I040-5Z  -f  o.oi69o8/2  -f  22.54417  -f  0.00252 1///2  -f  5.843/5  +  10.69-, 
e6  =  0.5202  5Z7  +  1040. 5^  -f  0.00751 5/2  -f  15.02907  -f-  O.OOI474///2  +  6.030/2  +  17.69-^, 
£8  —  0.390^7  +  1040.5^  -f  0.004227/2  +  1 1.27207  +  o.ooiO22/z72  -f  6.i84//  +  24.58-, 
fIO  =  o.3i2i5Z7+  1040.5^,  +  o.oo270572+  9.01767  +  0.0007  7  7  M 2  +  6.268^  +  31.43-, 
eI2  =  o.26oi3Z7  +  IO4O.5Z  +  o.ooiSjgt2  -f-  7.51477  -f-  o.ooo624//72  -)-  6.322/2  -f  38.26  *, 
fu  =  0.22297^7  +  1040.5^  +  o.ooi38o72  +  6.44127  +  0.000520//72  +  6.359/2  +  45.08-, 
«I6  —  0.19509^7  -f-  IO4O.5/,  +  O.OOIO5772  +  5.63607  +  o.ooo446//72  +  6.387^  +  51.90-, 
eI8  =  O.I7342Z7  +  1040. $L  -f  o.ooo83572  -(-  5.00987  -f  o.ooo389/^72  +  6.407/2  -f-  58.71-, 
e2Q  —  0.15608^7  -f-  1040. 5^  -f  o.ooo67672  +  4.50887  +  o.ooo346//72  +  6.424/5  -(-  65.53-, 
e22  —  o.i4i89Z7  +  IO4O-5Z  -f  o.ooo55972  -f  4.09897  +  0.00031 1//72  -f  6.437/2  -j-  72.34-, 
e24  =  o.i3oo6Z7  +  I040-5Z  +  o.ooo47o72  -f  3.75747  +  o.ooo282/272  +  6.448^  +  79.15-, 


which,  as  above  defined,  is  the  weight  upheld  vertically  by  each 
support. 

It  will  be  observed,  that,  in  all  terms  involving  hz  in  the 
expression  for  weight  of  vertical  supports,  equation  (522),  //2  has 
been  replaced  by  \hl.  This  substitution  is  simply  for  conven- 
ience, and,  being  made  in  these  small  terms  only,  does  not 
practically  affect  the  accuracy  of  our  resulting  equations,  while 
we  are  hereby  relieved  of  higher  powers  of  k  than  the  second, 
in  the  value  of  W. 

164.  As  additional  security  against  deflection,  out  of  the 
plane  of  the  girder,  by  the  top  chord,  we  shall  insert  a  system 
of  head  lateral  bracing  between  the  two  top  chords  where  the 


MECHANICS  OF   THE   GIRDER. 


height  is  sufficient.  For  this  purpose,  the  two  top  chords  are 
the  flanges  of  a  great  longitudinal  strut  or  column,  whose 
tendency  to  deflect  laterally  must  be  overcome  by  this  head 
web  system  of  diagonals  and  struts. 

Suppose  the  moment  at  the  centre  of  this  system  be  that 
due  to  \  W*  acting  upon  each  panel  length,  or  to  \  Wl  acting 
upon  the  windward  side  at  each  joint  of  the  windward  top 
chord  ;  that  is,  by  (480),  where  now  we  must  put  \n  for  ;/,  and 

\n  for  r,  and  2,500-  pounds  for  Wu  we  have 
n 

Moment  at  centre  =  M  =  -faWJn  =  78.125/^7; 
and  the  longitudinal  horizontal  strain, 

H  —  —  =  78.125—  pounds  at  centre, 
?i  ?i 

which  in  each  flange  may  be  considered  to  decrease  uniformly 
to  the  ends,  as.  is  practically  the  case  with  that  part   of   the 
strain  due  to  bending-moment  in  a  pillar. 
Therefore,  for  each  double  panel  length, 

*ff  =  H  X  ~  =  312.5—; 
\n  nq, 

requiring  each  diagonal  tie  to  resist 

312.5  /// 


and  to  have  a  cross-section 


312.5^7  . 

S  —  -  -  -  -  -  r  -  =  -  ^r  square  inches,       (523) 
15000  cos  a  cos  <f>2nql 


CALCULATION  OF  THE    WEIGHT  OF  BRIDGES.          457 

since  cos  a  may,  for  these  central  panels  receiving  the  head 
system,  be  put  =  i  without  practical  error. 

Length  of  each  head  diagonal  =   — practically. 

n  cos  <p2 

Weight  of  2! 3  )  wrought- iron  head  diagonals,  in  pounds, 


=  0.007716(7*  -  6)          + 


(524) 


=  h 


n 

-M/2 

•  - 

4 

o 

o 

6 

1.25 

0.0002411 

8 

2.50 

0.0003086 

10 

3-75 

0.0003215 

12 

5.00 

0.0003149 

14 

6.25 

0.0003014 

16 

7-5° 

0.0002857 

18 

8-75 

0.0002701 

20 

IO.OO 

0.0002551 

22 

11.25 

0.0002411 

24 

since 


ql  =  1 8  feet,     and 


COS2 


+  =  i  +  ta- 

\  2/  / 


If  we  multiply  5  in  (523)  by  2  X  1 0,000  cos  <£2  tan  </>2,  we 
have,  by  our  specifications,  the  end  pressure  brought  by  each 
pair  of  diagonals  upon  the  end  of  each  head  strut ;  and,  calling 
the  inch  strain  in  compression  on  these  struts  2,500  pounds, 
wo  have  the  cross-section  of  each  head  strut,  in  square  inches, 


453 


MECHANICS   OF   THE    GIRDER. 


312.5   X  2   X   ioooo///tan</>2 

15000  X   2500/2^,  "  1^    • 


h  being  in  feet,  and  tan  <£2  z=  ^. 


Weight  of 


I  -  —  2  J 


head  struts,  in  pounds, 


=  h 


n 

o 

4 

5 

6 

10 

8 

15  v 

10 

20 

12 

25 

14 

30 

16 

35 

18 

40 

20 

45 

22 

5° 

24 

(sa6) 


h  in  feet. 

Since  we  have  already  provided,  in  the  floor  system,  for  the 
whole  bending-force  of  the  wind,  and  are  now  simply  stiffening 
the  head  system  laterally  as  a  column,  or  to  meet  adjustment 
strains,  and  strains  due  to  imperfections  in  workmanship,  it 
will  manifestly  suffice  if  we  call  the  additional  chord  strain  in 
each  segment  of  top  chord  within  the  head  system  equal  to 


=  312.5 —  =  17.36$—  pounds. 


And,  as  6,400  pounds  is  the  allowed  inch  strain  in  top  chords, 
we  have 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


459 


Cross-section  to  be  added  to  each  segment  of  top  chord  due  to  strain 
on  head  diagonals,  square  inches, 


=  S  =  0.00271267 — . 


(527) 


Weight  of  added  iron  in  ( -    -  3  J  of  the  central  double  panels,  for  top 
chords,  in  pounds, 


(5=8) 


n 

- 

4 

0 

6 

0.0005651 

8 

0.0007234 

10 

0-0007535 

12 

0.0007381 

14 

0.0007064 

16 

0.0006698 

18 

0.0006329 

20 

0.0005978 

22 

0.0005651 

24 

165.  To  find  the  Necessary  Amount  of  Material  for  the 
Triangular  Web  System  of  Latticed  Struts  or  Columns. 

Let  /'  =  length  of  strut. 

d  =  effective  width  of  strut. 

A  =  area  of  both  flanges  in  section  normal  to  axis  of 

strut,  in  square  inches. 

Al  =  area  of  diagonals  in  the  same  section  normal  to 
axis  of  strut,  and  not  to  its  own  axis,  in  square 
inches. 
0  —  45°  —  inclination  of  diagonal  to  axis  of  strut. 

BI  —  allowed  inch  strain,  both  in  flanges  and  diagonals, 
in  the  present  case. 


460  MECHANICS  OF   THE   GIRDER. 

Then,  moment  at  centre, 

M  =  \AdB» 

M 

Longitudinal  flange  strain  at  centre  =  H  =  —  = 

a 


Now,  since  H  decreases  uniformly  from  the  centre  to  the 
ends,  at  least  practically, 


which  is  the  longitudinal  component  of  diagonal  strain. 
And 

s-  cos  0  —  L  =  strain  on  diagonal ; 

/  COS0 


Ad  ,       v 

—*  (529) 


=  A  if//-^-  ^  =  20> 
=  TVif/A-^  =  30, 

=     JQ     If  /'    -5-     </     =     40, 


Since  about  one-half  of  each  diagonal  bar  is  cut  away  to 
receive  its  end  pin,  we  have  for  use, 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          461 


=  -       if/'-s-  d  =  20, 

5 

=  _I     if/'  -*-</  =  30, 

=  -L     if  /'  -*-  d  =  40, 
10 


I2-5 


=  50, 


-  JL     if/'.,.  ^  =  60, 


which  is  the  ratio  of  the  section  of  the  diagonal  bar  to  that  of 
the  two  flanges,  the  section  being  normal  to  axis  of  the  strut 
in  both  cases. 

This  ratio  must  be  doubled  for  square  struts  latticed  against 
deflection  both  ways,  and  it  becomes 

A,       M  ,      v 

~A  =  ~7'  (532) 

-   _L      if  /'  ^.   d   =    20, 

2-5 

=  _L  if  /'  _,_  d  =  30, 

3-75 

=  -        if  /'  -*-  d  =  40, 


=  JL    if  /'  _s_  ^  =  60. 
7-5 

By  reviewing  our  compression  members,  which  are  to  be 
latticed  in  at  least  one  direction,  we  find  the  girder  diagonals 


462  MECHANICS  OF  THE   GIRDER. 

having  the  ratio  of  length  to  radius  of  gyration  =  100,  giving 
ratio  of  length  to  diameter  =  about  40:  so  that,  by  (531),  the 
weight  found  in  (491)  should  be  augmented  by  one-tenth  of 
itself.  Also,  the  vertical  supports  have  a  mean  ratio  of  length 
to  width  =  2  X  10  =  20:  so  that,  by  (531),  that  part  of  their 
weight  due  to  bending-moment,  (521),  should  be  augmented  by 
one-fifth  ;  or,  which  is  approximately  the  same  thing,  the  weight 
given  in  (522)  is  to  be  increased  by  one-tenth  of  itself.  Simi- 
larly, we  shall  augment  the  weight  of  the  lateral  head  struts, 
(526),  by  one-tenth  of  itself,  on  account  of  bracing. 

In  general,  the  longitudinal  wind  chords,  being  attached  to 
the  floor  joists,  to  the  transverse  I-beams,  and  to  the  girder 
diagonals,  will  need  diagonal  bracing  only  when  very  long. 

The  top  and  bottom  chords,  however,  though  not  having 
diagonal  bracing  in  themselves,  yet  will  need  to  have  their 
weight,  (484),  (485),  augmented  by  about  one-tenth  of  itself,  on 
account  of  the  enlarged  ends  of  I-bars,  the  re-enforcement  of 
plates  and  rivets  at  joints,  and  the  nuts  and  pins. 

166.  Weight  of  the  Bridge.  —  Increasing,  therefore,  by 
one-tenth  of  itself,  the  weight  of  girders,  of  vertical  supports, 
and  of  lateral  head  struts,  and  collecting  all  the  weights  which 
will  then  have  been  found  in  pounds,  and  expressed  in  terms  of 
W,  L,  /,  and  /i,  for  each  va^ue  of  n,  the  number  of  panels,  and 
putting  each  sum  =  2000/2  W  —  total  weight  of  bridge,  in 
pounds  also,  since  W,  the  panel  weight  of  bridge,  and  L,  the 
panel  weight  of  uniform  discontinuous  live  load,  are  in  tons,  we 
find  the  following  values  of  W  for  the  different  values  of  ;/, 
remembering  that  h  and  /  are  in  feet : 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


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466  MECHANICS  OF   THE   GIRDER. 

In  all  these  cases,  the  value  of  h  which  renders  W  a  mini- 
mum has  been  found  by  the  simple  method  of  article  140,  equa- 
tions (469)  and  (470).  The  limiting  spans  just  given  have  been 
determined  by  putting  the  denominator  of  (543)  equal  to  zero, 
and  substituting  the  assigned  values  of  h.  It  will  be  seen  that 
these  limiting  spans  are  independent  of  the  live  load,  nL,  and 
therefore  represent  the  limit  to  the  length  of  each  girder 
imposed  by  its  own  weight.  The  effect  of  live  load  on  the 
limiting  span  will  be  considered  below. 

167.  Having  found  W  and  //,  it  is  easy  to  compute  the  weights 
of  all  parts  of  the  bridge  from  the  expressions  for  weights  in 
terms  of  W,  /i,  L,  and  /.  The  computation  affords  a  perfect 
verification  of  the  accuracy  of  the  work.  We  give  below  a  table 
showing  the  number  of  panels  and  the  height,  which  simultane- 
ously render  the  total  bridge  weight,  nW,  a  minimum  for  various 
spans  ranging  from  50  to  1,000  feet,  and  have  thus  probably  ex- 
tended the  table  far  beyond  any  economical  use  of  this  girder. 

Of  course,  we  find  great  heights  ;  but  it  should  be  remem- 
b**ed  that  one-half  of  this  central  height,  h,  is  below  the  plane 
of  the  floor  system,  where  the  points  of  support  are  situated. 
Also  the  width,  18  feet,  becomes  too  small  for  the  highest 
girders  ;  but  it  has  been  retained  in  this  set  of  examples,  to  pre- 
serve uniformity  in  data.  » 

To  illustrate  the  change  in  central  height  and  bridge  weight, 
as  the  number  of  panels  varies  for  the  same  span,  we  have  given 
the  solutions  corresponding  to  three  values  of  n>  including  that 
one  which  renders  n  W  least.  Also  bridge  weights  and  central 
heights  are  given  for  2,000,  3,000,  and  4,000  pounds  of  live  load 
to  the  running-foot ;  the  weights  being  minima  values.  The  3Oth 
line  of  this  table  exhibits  the  effect  of  a  small  live  load  upon  the 
length  of  the  limiting  span,  as  resulting  from  the  substitution  of 
\W  for  L  in  the  equations  for  weight.  Of  course,  we  do  not  mean 
that  the  live  load  is  small  near  the  limit  when  W  is  infinite. 

The  reader  cannot  fail  to  notice  how  prolific  in  useful  and  in- 
teresting results  these  general  equations  for  bridge  weight  are. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


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CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          4/1 

168.  Among  the  inferences  which  may  be  legitimately 
drawn  from  the  table  of  article  167  in  regard  to  bridges  having 
two  lenticular  Brunei  girders  of  single  system  of  the  same 
width  but  of  varying  span,  height,  and  uniform  live  load,  are 
the  following  ;  viz.  (see  Fig.  116),  — 

ist,  The  best  number  of  panels  varies  approximately  as  the 
cube  root  of  the  span, 

n  <*  fi  nearly.  (544) 

2d,  The  best  central  height  for  a  uniform  live  load  of  2,000 

pounds  per  linear  foot  is  about  —  X  span, 

4.2 

h  =—l  nearly;  (545) 


and  for  different  loads  the  best  height  for  same  span  and  same 
number  of  panels  varies  very  nearly  as  the  sixth  root  of  the 
live  load, 

h  oc  (nL)~*  nearly.  (546) 

3d,  For  spans  less  than  500  feet,  the  least  bridge  weight 
varies  approximately  with  the  product  of  the  best  central 
height  of  girder  multiplied  by  the  span  (that  is,  with  the  geo- 
metrical area  of  the  girder,  since  the  parabolic  area  is  propor- 
tional to  this  product)  ; 

/  <  500,     nW  c<  lh  nearly.  (54?) 

4th,  For  the  same  span  and  same  number  of  panels,  and 
best  central  height  of  girder,  the  least  bridge  weight  varies 
approximately  with  the  square  root  of  the  uniform  live  load  ; 

nW  'oc  (nZ,y*  nearly.  (548) 

Or,  by  reason  of  (546),  the  least  bridge  weight  for  same  span 


472  MECHANICS  OF   THE   GIRDER. 

and  best  central  height  of  girder  varies  nearly  as  the  cube  of 
this  best  central  height ; 

nW  oc  (nL)*  oc  fa  nearly.  (549) 

5th,  For  each  span  of  500  feet  and  over,  a  large  increase  of 
live  load,  and  consequently  of  best  height,  causes  a  diminution 
in  the  number  of  panels  corresponding  to  minimum  bridge 
weight. 

6th,  Small  deviations  from  the  best  height  and  best  number 
of  panels,  or  from  either  of  them,  do  not  greatly  affect  the 
bridge  weight ;  but  large  deviations  either  way,  in  this  respect, 
cause  a  great  increase  in  bridge  weight,  as  shown  in  sixth  line 
of  table,  h  =  -j1^/,  thus  rendering  the  girders  only  about  one-half 
as  high  as  minimum  bridge  weight  requires. 

7th,   The  limiting  span  increases  slowly  with  the  number 

of  panels,  till  a  maximum  value  depending  upon  -—  and  -  is 

JLf  It 

reached. 

8th,  We  cannot,  for  a  given  span,  assign  the  best  height 
and  the  best  number  of  panels  till  we  know  the  live  load  which 
is  to  be  imposed. 

169.  EXAMPLE.  —  Take  span  /  =  200  feet,  number  of  panels 
n  =  14,  central  height  of  double  parabola  h  =  42.585  feet, 
uniform  live  load  nL  =  200  tons  =  400,000  pounds,  width  of 
2|~inch  oak  floor  q  —  16  feet,  length  of  transverse  I  floor 
beams  ^=18  feet.  (See  Fig.  16.)  Loads  applied  at  all 
apices  equally  by  means  of  struts  and  suspenders  which  sustain 
the  floor  system  in  the  plane  of  the  axes  of  girders. 

Assume  that  all  cross-sections  of  members  may  be  strictly 
adjusted  to  the  developed  strains. 

Weight  of  live  load 400000  pounds, 

Weight  of  floor,  article  167     .     .     .     .       34667  pounds. 

Total  load  on  longitudinal  I-beams  .     .     434667  pounds. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          4/3 

Load  on  i  panel  length  of  each  longitudinal  I-beam  spaced  3.2  feet 
434667 


By  (502), 

Cross-section  of  beam  =  S  =  o.oooisf  ¥?  X  —  +  2000  X  — 

\  3 

=  4.65714  square  inches. 

In  order  to  satisfy  the  condition  in  (501),  we  must  have, 
article  62, 

(550) 


ion 

S  =  bd  -  b^  =  4.65714;  (551) 

from  which  equations  we  find 
O  =  (#  _  bb*)d*  -  $b2Sd2  -f    tfS*  +  i.2-^/  -  S3.      (552) 


Take  £  =  4.00  inches  =  breadth  of  flange. 
b  —  b,_  =  0.26  inch  =  thickness  of  web. 
Then,  from  (552)  and  (551), 

d  =     9.080  inches  =  depth  of  beam, 
d  —  </!  =     0.614  inches  =  depth  of  two  flanges, 
/  =  60.466  =  moment  of  inertia, 

7  =  I  x  lli  x      62°9-53     =  60.466  =  6>6 
^/        8          «          2   X   10000          9.08 

by  reason  of  (502)  and  (52)  ;  the  load  being  uniformly  distrib- 
uted on  each  panel  length  of  beam,  and  these  beams  not  being 
regarded  continuous  over  the  transverse  beams. 

Weight  of  these  6  longitudinal  I-beams,  by  (503),  equals 

/,  =  6  x  fg  x  12  x  200  x  4.65714  =  18628  pounds, 
as  ^ivcn  in  preceding  table. 


474  MECHANICS   OF   THE   GIRDER. 

It  will  be  noticed  that  these  beams  are  deep  and  compara- 
tively thin  ;  but,  considering  their  area  of  cross-section,  it  will 
also  be  noticed  that  their  moment  of  inertia  is  great  as  com- 
pared with  ordinary  beams  of  equal  area  of  section. 

Supported  by  the  transverse  I-beams,  we  have 

Live  load  =  400000  pounds, 

Floor  =     34667  pounds, 

Longitudinal  I-beams  =     18628  pounds. 

Total  for  14  panels        =  453295  pounds. 
Load  on  i  beam  =    32378  pounds. 

From  (504),  we  have 

/       12  x  18  X  32378 

d  =      8  x  2  x  10000    =  4 

by  (505)  ; 

.*.     S  =  21.855  square  inches  for  vertical  load. 

But,  in  order  to  resist  the  assumed  wind  pressure,  W^  = 
2,500-  —  7,604  pounds  per  panel,  we  must  add  to  the  cross- 
section  due  vertical  load  the  areas  found  from  (511),  where  now 

O2  =  -  —  —  -   =71500  pounds  per  square  inch  ; 
I  +  0.93312   X  0.07 

2    X     7604. 


3  x  .4  x 

=  8.149  square  inches,  ist  and  i3th  beams; 
=  6.944  square  inches,  2d  and  i2th  beams; 
=  5.835  square  inches,  3d  and  nth  beams; 
=  4.822  square  inches,  4th  and  loth  beams; 
=  3.906  square  inches,  5th  and  9th  beams; 
=  3.086  square  inches,  6th  and  8th  beams; 
=  2.363  square  inches,  7th  beam. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          4/5 

TOTAL  SECTIONS. 

S  +  Si  =  30.004  square  inches,  ist  and  i3th  beams; 
28.799  square  inches,  2d  and  i2th  beams; 
27.690  square  inches,  3d  and  nth  beams; 
26.677  square  inches,  4th  and  roth  beams; 
25.761  square  inches,  5th  and  9th  beams; 
24.941  square  inches,  6th  and  8th  beams  ; 
24.218  square  inches,  7th  beam. 

351.962  =  2<S  =  sum  of  all  sections. 
To  satisfy  the  condition,  (505),  we  must  now  have 

/  =  -^(bd*  _  ^3)  =  2&/  =  2d(dd  -  <V/0;     (553) 
whence,  eliminating  </„  we  find 

O  =  b(b*  -  b*}d*  -  s^Sd2  +  (3^2  +  2^l2S)d  -  S3,     (554) 


from  which  d  may  be  found  for  each  value  of  total  section  now 
called  S\ 

bd  -  S 


Taking  b  =    5.5  inches  =  width  of  flange, 
b  —  #x  =    0.7  inch      =  thickness  of  web, 
bl  =.    4.8  inches  =  difference, 
S  =  30.0  square  inches  =  cross-section, 
we  find,  by  (554)  and  (553), 

d  =  13.441  inches  =  depth  of  beam, 
dl  =     9.152  inches  =  depth  of  web, 
d  —  dl  =     4.289  inches  =  depth  of  both  flanges, 
-J(//  —  d^)  =     2.144  inches  =  depth  of  one  flange, 

/  =  806,     -  =  60. 
d 

Similarly  may  the  proportions  of  the  other  transverse  beams 
be  found. 


4/6  MECHANICS  OF   THE   GIRDER. 

Or,  if  we  choose  to  assume  the  thickness  of  web  and  of 
flanges,  thus  : 


d  -d,  =  c  (say), 
then  we  find,  from  (553), 

O  =  <t*  -  |(^  +  A/2  +  |  (12  +  fO-  +  ~\<l  ~  —  ,      (556) 
\a         /  (  a        2  }  za 

from  which  d  is  easily  found  either  by  trial  or  by  Horner's 
Method. 

Taking  a  ==.  0.7,  c  =  4,  5  =  30,  we  find,  by  (556), 

d  =  13.2,         /.     d,  =  9.2. 

But  J  =  S  +  a  -  ^,  by  (553)  and  (555), 
£*  ^ 

=  5.889; 

/.     b,  =  5.189, 

7  =  792'    ^  =  6°* 
Or  again,  by  assigning  values  to  d  and  di  in  (553),  we  find 

=  (24  -  d)dS  (       } 

di(d*  -  d^ 

b  =  ^24  ~  ^)^  +  ^.  (558) 

,/2     _     d*  r    d 

Using  two  1  2-inch  beams  for  each  panel  point,  we  have 
d  =  12,     di  =  10,     d  —  dt  =  2, 

b  =  5.3416          ^r  =  4.9098         b  —  bl  =  0.4318  inch. 

=  5.1272  =  4.7127  =  0.4145  inch. 

=  4.9296  =  4.5311  =  0.3985  inch. 

=  4.7490  =  4.3652  =  0.3838  inch. 

=  4.5864  =  4.2156  =  0.3708  inch. 

=  4.4404  =  4.0814  =  0.3590  inch. 

=  4.3116  =  3.9629  =  0.3487  inch. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          477 

Whatever  be  the  form  of  beam  section  chosen,  we  have 
Weight  of  13  X  2  transverse  I-beams 

=  12  x  18  x  T5s^  =  60  x  351.962  =  21118  pounds, 
as  per  table. 

Strains  on  the    horizontal    diagonals   are   given   by  (508), 
where  now  W,  —  7,604,  and  sin  </>x  —  0.78329; 


2n  X   15000  sin  (^ 


0.023114. 


0.023114  x  14  X  13  =  4.206  square  inches  =  section  of  ist  diagonal, 

0.023114  x  13  X  12  =  3.605  square  inches  =  section  of  2d  diagonal, 

0.023114  x  12  x  IT  =  3.051  square  inches  =  section  of  3d  diagonal, 

0.023114  x  ii  X  10  =  2.543  square  inches  =  section  of  4th  diagonal, 

0.023114  x  10  x     9  =  2.080  square  inches  =  section  of  5th  diagonal, 

0.023114  x  9  X     8  =  1.664  square  inches  =  section  of  6th  diagonal, 

0.023114  X  8  x     7  =  1.295  square  inches  =  section  of  7th  diagonal. 


18.444  X  4  =  3S  =  73.776  square  inches. 

=  12  x  —  X  - 
1  8       s 

=  5652  pounds, 


Weight  of  28  horizontal  diagonals  =  12  x  —  X  -7^  —  X  73.776 

1  8       sin  <x 


as  given  in  the  table. 

The  cross-section  of  each  panel  length  of  a  wind  chord  is 
shown  in  (514),  thus  : 

J*V_  =  _  7604  X   200  _ 
**4iQ      2  x  14  x  1  8  x  6400 

0.471478  x  13  =  6.129  square  inches,  ist  panel; 
O.47I478-X  24  =  11.316  square  inches,  2d  panel; 
0.471478  x  33  =  15.559  square  inches,  3d  panel; 
0.471478  x  40  =  18.859  square  inches,  4th  panel; 
0.471478  x  45  =  21.217  square  inches,  5th  panel; 
0.471478  x  48  =  22.631  square  inches,  6th  panel  ; 
0.471478  x  49  =  23.103  square  inches,  7th  panel. 

Total,  118.814  square  inches  for  one-half  of  i  girder. 


.47$  MECHANICS   OF   THE   GIRDER. 

.-.    Weight  of  both  wind  chords  =  4  X  -5-  X  I2  X  2°°  x  118.814 

18  14 

i 

=  22631  pounds, 

as  by  (515). 

We  now  have,  upon  all  vertical  supports  and  abutments, 

Live  load      .....  400000  pounds, 

Floor  .......  34667  pounds, 

Longitudinal  I-beams     .  18628  pounds, 

Horizontal  diagonals  .     .  5652  pounds, 

Wind  chords     .     .     .     .  22631  pounds. 

•£%  x  481578  pounds  =  17199  pounds, 

H  --  X  weight  of  transverse  I-beams  =  -      -pounds  =       812  pounds. 
26  26 

Load  on  each  vertical,  article  163,  =  sn  =18011  pounds. 

Therefore,  by  (516), 

,5  =  %%¥"  =  3-3771  square  inches  =  cross-section  of  a  strut, 
due  vertical  forces  ;  and,  by  (518), 

(ToV"  =  3«ooi8  square  inches  =  cross-section  of  a  suspender, 


due  vertical  forces. 
From  (520), 

75        200 
S2  =  -j—^  x  ~^~  —  6.3025  square  inches  =  cross-section  of  each  vertical 

due  bending-moment  of  assumed  wind  force  ; 

/.     S  +  S2  =  9.6796  square  inches  for  each  strut, 

S,  -}-  S2  =  9.3043  square  inches  for  each  suspender. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


479 


From  (473),  we  have  length  of  verticals, 

Suspenders. 

Struts. 

y 

=  0.43404  x   13  =       5.6425  feet. 

r  =  i 

0.43404  x  24  = 

10.4170  feet. 

2 

0.43404  x  33  =     14-3233  feet. 

3 

, 

0.43404   X   40  = 

17.3616  feet. 

4 

0.43404  X  45  =     T9-53l8  feet. 

5 

0.43404   X   48  = 

20.8340  feet. 

6 

0.43404  x  49  =     21.2680  feet. 

7 

Sum  required  =  200.5268  feet. 

194.4504  feet,  for 

all. 

Longest  strut,  20.834  feet  =  250  inches; 
therefore 

Required  radius  of  gyration  =  -fgj}  =  2\  inches. 

Each  vertical  may  be  made  of  4  channels,  6  inches  wide, 
each  having  an  area  of 

2.4199  square  inches  for  struts, 
2.3261  square  inches  for  suspenders, 

latticed  in  pairs,  and  two  pairs  in  one  brace. 

Weight  of  all  vertical  struts 

=  -fg  x  12  x  194.4504  x  9.679  =     6273  pounds, 
Weight  of  all  vertical  suspenders 

=  -fs  x  12  x  200.5268  x  9.304  =     6218  pounds. 

Total,  12491  pounds. 

Add  one-tenth  for  lattice  braces,  1 249  pounds. 

13740  pounds, 
which  accords  with  (522). 

Equation  (523)  gives  the  cross-section  of  each  head  diagonal 
thus  : 

s  =  O.Q2081  X  42.585   X   200  =  square  inch) 

14  x   18  X  0.84609 

which  requires  a  round  rod  1.056  inches  in  diameter  if  the  ends 
are  enlarged  for  cutting  threads  of  screws. 


480  MECHANICS   OF   THE    GIRDER. 

Weight  of  the  8  head  diagonals,  by  (524),  is 

8  x  ^  X  I2  X  2  *  20°  X  0.831  =  749  pounds. 
18         14  x  0.84609 

Cross-section  of  each  head  strut,  by  (525),  is 

yV  X  42-585  =  3-549  square  inches. 
Add  one-tenth  for  latticing,  0.355  square  inch. 

3.904  square  inches. 

Weight  of  5  head  struts,  by  (526)  =  25  x  42.585  =  1065  pounds. 
Add  one-tenth  for  braces  =  106  pounds. 

Total,  1171  pounds. 

Since  for  these  head  struts  we  have  assumed 

8000 
Q  =  2500  =  -  -  —  pounds  per  square  inch, 


2OOOO 

12  X   18 


OC  —  —    2IO  ^-^          —  — 


p  P 

.•.     p  =  |ij  =  1.03  inches  =  radius  of  gyration. 

We  may  therefore  use,  for  each  head  strut,  2  4-inch  channels 
latticed  so  that  the  web  shall  be  4  inches  apart. 

The  increment  of  section  of  each  top  chord  due  to  diagonal 
strain  in  head  system  is  given  by  (527),  thus  : 

42.585  x  200 

S  =  0.00271267  X  -  =  1.65  square  inches. 

*4 

The  total  weight  thus  added  along  the  4  panel  lengths  of 
head  system  is 

2  X  4  X  -5-  X   I2   X   2  X   20°  x   !.65  =  1257  pounds, 
18  14 

as  by  (528). 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          48  I 

The  strain  in  the  top  chords  is  given  by  equations  (476), 
(481),  and  (482),  where  now 

W  -f-  L  =  7.72265  -f-  14.28571  =  22.00836  tons. 

For  the  segments  of  each  top  chord,  the  total  strains  due 
n(  W  +  L)  are 

Pt  =  101.15  tons> 

^2     =          96-55     t011S> 

/>  =     93.10  tons, 
P4  =     91.55  tons. 

Dividing  these  strains  by  the  allowed  inch  strain,  Q  =  3.2 
tons,  we  get  cross-sections, 

5,  =  31.6075  square  inches  •+•  ) 

S,  =  30.  1  704  square  inches  }  for  head  system> 

S3  =  29.0937  square  inches  -f-  1.65  square  inches, 

*S4  =  28.6078  square  inches  +  1.65  square  inches. 

Now,  the  longest  unsupported  segment  of  top  chord  is 

2/ 

'-  —  =  29.405  feet  =  352.86  inches. 
n  cos  «2 

/.     3.5286  inches  =  radius  of  gyration. 

Therefore  the  top  chord  may  be  made  up  of  2  Q-inch  chan- 
nels and  a  plate,  or  2  plates  14  inches  wide,  and  having  such 
thickness  as  is  required  to  complete  the  area  of  section. 


Weight  of  top  chords  due  load 


=  —  X  —  X  -^—2—^  —  =  44741  pounds, 
10        18          n     cos  a 

1257  pounds,  due  head  system. 
Total,  45998  pounds. 


482  MECHANICS  OF   THE   GIRDER. 

Similarly,  for  the  segments  of  each  bottom  chord,  the  total 
strains  due  n(W  +  L)  are,  from  equations  (477),  (481),  and 

4483), 

£/i  =  100.128  tons, 

U2  =  99.513  tons, 

Uz  =  92.133  tons, 

^4  =  9 1  -3  74  tons. 

These  strains  divided  by  5,  the  allowed  inch  strain  in  ten- 
sion, give  the  cross-sections  of  the  successive  segments  of 
bottom  chord  in  each  girder, 

St  =  20.0256  square  inches, 
S2  =  18.9025  square  inches, 
S5  =  18.4265  square  inches, 
S4  =  18.2748  square  inches, 

from  which  the  links  can  easily  be  made  up  according  to  speci- 
fied forms  of  body  and  head. 

No  change  is  here  made  on  account  of  longitudinal  compo- 
nent of  lateral  diagonal  strain,  since  in  the  present  case  there  is 
no  lateral  system  between  bottom  chords,  by  reason  of  gravity. 

Weight  of  all  bottom  chords  increased  by  -^ 

=  —  X  -5-  x  ^2  —  =  28695  pounds. 
10        18         11      cos/? 

The  equations  (490),  (491),  and  (478)  give  cross-sections  of 
alternate  girder  diagonals,  thus  : 

St  =  5.115  square  inches, 

52  =  7.212  square  inches, 

53  =  8.664  square  inches, 

54  =•  8. 88 1  square  inches, 

55  =  7.750  square  inches, 

56  =  5.132  square  inches, 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


483 


for  each  of  the  two  girders,  the  alternate  set  being  the  same 
inverted  ;  and  the  weight  of  all  is,  calling  <2i  —  f  >  and  multi- 
plying by  1.8  X  -i"J,  as  specified, 

4  x  J_  x  3  X  1.8  X   ii  x  12  x  200^    S     _  990^   S 
1 8  8  x   10  14          cos0         7     cos# 


=  23184  pounds. 

Now,  since  the  longest  unsupported  length  of  any  girder 
diagonal  is 

-  X  =  22.253  f£et  =267  inches. 

2        n  cos  0A 


we  have  radius  of  gyration  =  2.67  inches ;  and  therefore  2 
8-inch  channels  latticed  8  inches  between  webs  will  suffice 
for  the  longest  diagonals. 

We  have  thus  determined  the  size  and  weight  of  all  parts  of 
this  bridge,  and  find  the  total  216,233  pounds,  as  by  table. 

STRAIN  SHEET. 

STRAINS,  TONS;    CROSS-SECTIONS,  SQUARE  INCHES. 
For  each  of  Two  Girders. 


20. 


FIG.  118. 

Span,  200  feet;   central  height,  42.585  feet  (best) ;   uniform  live  load,  i  ton  =  2,000  pounds  per  linear 
foot,  applied  at  all  apices  by  vertical  members. 

Regarding  the  greatest  strains  upon  the  chord  pins  as  acting 
in  "quadruple  shear,"  and  allowing  6,000  pounds  as  the  inch 
strain  in  shearing,  these  pins  will* require  a  diameter  of  3^ 
inches. 


484  MECHANICS   OF   THE   GIRDER. 

It  remains  to  compute  the  deflection  of  this  girder  under 
the  allowed  chord  strain  of 

BI  =  -J(3-2  -f  5)  =  4.1  tons  per  square  inch. 

For  this  purpose,  use  equations  (318)  and  (319)  combined 
thus  : 

D  =  fr ^-3862950  -  2.302585[0  +  *)log(a  +  x) 
jf£At 

+  (a  -  x)log(a  —  x)  -  2a  log  a]  \,     (559) 

where  a  =  J/  =  100  feet  =  half-span,  and  x  is  measured  from 
the   centre.     Also   /^  =  h  —  42.585    feet,   and  we   will   call 
E  =  24,000,000  pounds  =  12,000  tons  per  square  inch. 
We  have  then,  from  (559), 

Deflection  Dl  =  1.3347  inches  for  x  =    o,  centre  ; 

100 
J}2  =  1.3150  inches  for  x  =  » 

200 
D3  =  1.2549  inches  for  x  =  ——  > 

300 
/?4  =  1.1521  inches  for  x  =  — > 

400 
£>5  =  i  .0006  inches  for  x  =  —  > 

500  < 
D6  •=.  0.7911  inch     for  x  =  — > 

600 
D7  =  0.4954  inch     for  x  —  ——  > 

A  =  o  inch      for  x  =  — »  ends- 

The  proper  camber  may  be  given  to  the  girders  by  equation 
(366),  thus  : 

X  =  — X  length  of  one  parabolic  chord. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          485 

This  length  is  given  by  equation  (140) ;  viz., 

=  £(2oo2  +  4  x  42-5852)* 

2002         2h  -f  (/2  +  4/;2)* 
+  °'287823  X  log-      -"       ~  =  2°5'5  feet> 


.-.     A  =  -  -   x  205.5  =  1-685  inches 

12000 

=  length  to  be  added  to  top  chord. 
Or,  each  segment  should  be  lengthened  by 

-~- —  —  0.241  =  J  inch  nearly. 

SECTION  3. 
The  Brunei  Girder  of  Double  System. 

170.  We  now  take  the  girder  shown  in  Fig.  22  ;  but  we  will 
apply  the  dead  and  live  loads  at  all  apices  by  means  of  verticals 
whose  upper  half  will  act  in  tension,  and  lower  half  in  com- 
pression. These  verticals  must  also  resist  bending-moment 
due  wind.  Each  girder  has  two  equal  parabolic  chords,  and  the 
floor  system  is  in  the  plane  of  girders'  axes  ;  each  panel  length 
of  chord  is  straight,  and  the  number  of  panels  may  be  odd  or 
even ;  each  system  will  be  assumed  to  do  one-half  of  the  work. 

The  height  between  the  two  parabolic  arcs  at  the  centre 
being  /i,  the  height  at  any  apex  is  given  by  (474).  Equation 
(473)  gives  y. 

tan«r    =  —  tan  ft  =  — -      ^  r       -  =  ~r(n  —  r  —  r-*)>        (56°) 
cos2ar       cos2  ft  n2l2 


486  MECHANICS   OF   THE   GIRDER. 

,  n   _     (  \n 

I 

I  I  .          A./1      r-        f  >..  x  \  ~1  2  //T\ 

171.  Moments  at  all  apices  due  total  dead  and  live  uniform 
loads  are  given  by  (65), 

jf  -W  +  L,  . 

2Vllf       —  f(7l /    J    I       y 

2/2 

and  the  horizontal  component  of  chord  strain  is 

-Z)^;  (564) 


that  is,  this  component  is  uniform  throughout  the  girder  under 
uniform  load. 

TT 

Strain  in  top  chords  =  P  =  , 

COStt 

TT 

Strain  in  bottom  chords  =  U  =  , 

COSytf 

Cross-section  of  top  chords         =  P  -i-  Q,     ^=3.7647; 
Cross-section  of  bottom  chords  =  U  H-  Tt     T  =  5.0000. 

Volume  of  a  segment  of  top  chord  is,  therefore, 

*2lH      =  UW  +  Z) cubic  inches.  (565) 

nQcos2a  JQ/icos2a 

Weight  of  top  chords,  in  pounds, 


4g      _  i\         (     6) 
r  3^2\         »/ 

+  0.14757^  ~  ^V  | 


2  x 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


487 


W  4-  L 

n 

'       \    j-t 

0.442709/2  4-  0.55338^ 

4 

h 

/*• 

°-553386/2  4-  0.70833/^2 

5 

0.664063^  4-  o.86o82/z2 

6 

0.774740/2  4-  I.OII9O/*2 

7 

0.885418/2  4-  1.16211/^2 

8 

0.996095/2  4-  1.31172/^2 

9 

i.io6772/2  4-  1.46093^ 

10 

1.217449/2  4-  i.  60984^ 

ii 

i-328i26/2  4-  i.  75853^ 

12 

i.4388o4/2  4-  1.90705^ 

13 

1.549481/2  4-  2.05542/^2 

14 

1.660158/2  4-  2.20370/42 

15 

1.770835/2  4-  2.35188/^2 

16 

1.881512/2  4-  2.50000^ 

17 

1.992190/2  4-  2.64804/^2 

18 

2.102867/2  4-  2.79610/^2 

19 

2.213544/2  4-  2.94400/1* 

20 

2.324221/2  4-  3.09192/^2 

21 

2.434898/2  +  3.2398^2 

22 

2-545576/2  +  3.38767^ 

23 

2.656253/2  4-  3.53554^ 

24 

Similarly,  we  have 

Weight  of  bottom  chords,  in  pounds, 
yn(W  4-  Z)/2X       2/a       (IV  -\-L}l2(      ,   4//2/          i\  )    ,   ,   , 

W  4-  L 

A     */  ) 

__     **          \^    •L' 

~h 

Q-333333/2  +  0.41667/^2 

4 

/*• 

0.416667/2  4-  0.53333/^2 

5 

0.500000/2  4-  o.  648  1  5/;2 

6 

°-583333/2  +  0.76191/^2 

7 

0.666667/2  4-  0.87500^ 

8 

0.750000/2  4-  0.98765/^2 

9 

°-833333/2  +  i.ioooo/^2 

10 

MECHANICS  OF   THE   GIRDER. 


W  +  L 


i.oooooo/2  -f-  1.32407^ 

i-o83333/a  4-  i.43590/*2 

i.i6666y/2  4-  1.54762^2 

i. 250000/2  -|-  1.65926/^2 

1-333333^  +  I.77Q83/*2 

1.416667/2  4-  I.88235//2 

I.500000/2  4-  I.99383/*2 

i-583333/2  +  2.10526/^2 

I.66666;/2  -1-  2.21667/^2 

I.75OOOO/2  4-  2.32804^ 

I-833333/2  +  2.43939/^2 

i.9i6667/2  4-  2.55072/z2< 

2.OOOOOO/2  4-  2.66204/^2 


II 

12 

13 
14 

J5 
16 

i7 

18 

J9 

20 

21 

22 

23 

24 


172.  For  the  advancing  uniform  live  load  of  \L  at  each 
upper  and  lower  apex,  or  of  L  at  each  vertical  section  through 
apices,  we  have  at  foremost  end,  by  (64)  and  (474), 


z/, 

2?z2 


»  -  r) 


^ 

n  —  r) 

' 


=       (r  + 


(568) 


and  at  one  interval  before  the  foremost  end  of  live  load,  by  (68) 
and  (474), 

~r(r  +  i)(«  -  r  -  i) 
(HL)^  T  =  ^t-1  -  ^7 =  ^r.     (569) 


—  (?*  4-  i)  (»  —  r  —  i) 

/z2 


Therefore 


(570) 


which  is  the  horizontal  component  of  strain  on  both  diagonals 
of  a  panel,  on  the  present  assumption  that  the  two  diagonals  do 
equal  work,  and  that  the  whole  load  is  on  one  girder. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


489 


Hence,  for  each  of  two  girders,  we  shall  have 


Cross-section  of  a  girder  diagonal  = 


t  cos 


(571) 


according  to  our  specifications  for  members  alternately  in  com 
pression  and  tension.     Putting  m  =  -f$,  <2X  —  f,  we  find 

Weight  of  girder  diagonals,  pounds,  =  4  X  \zrnl  X  °- 

onh  Qi 


!T f-   _4 — 
n        i5/2 


n2 


L 

h 

0.140625/2  4-     i-722667z2 

4 

fi 

o.i6875o/2  4-     3.09600/52 

5 

0.187500/2  4"    4'77°83/^2 

6 

0.200892/2  4-     6.743447^ 

7 

0.210937/2  4-    9-oi3i8^2 

8 

0.218750/2  4-  11.58025^2 

9 

0.225000/2  4-  14.445007^ 

10 

0.230113/2  4-  17.607817^ 

ii 

0.234375/2  -f-  2i.o69Oi7z2 

12 

0.237980/2  4-  24.828S67;2 

T3 

0.241071/2  4-  28.88759^ 

14 

0.243750/2  4-  33.245337^ 

15 

0.246093/2  4-  37-90228^2 

16 

o.248i6i/2  4-  42.858547^ 

17 

0.250000/2  4-  48.11420/^2 

18 

0.25  1644/2  4-  53-66934/J2 

X9 

o.253i25/2  -f  59.524037^ 

20 

0.254463/2  4-  65.678337^ 

21 

0.255682/2  4-  72.I3228/*2 

22 

0.256803/2  -h  78.885927^2 

23 

0.257812/2  4-  S5.93I79/*2 

24 

L\  /         2\  /  30        63        30 

=  -r<  0.28 1 25!  i I/2  4-  o.o7s(  2/z2  —  5  —  —  H — -2 ; 

h  (  D\         ;z/  /:>V  n        n         nz 


490 


MECHANICS   OF   THE    GIRDER. 


All  girder  diagonals  must  be  so  constructed  as  to  transmit 
stresses  of  tension  or  compression. 

173.  Collecting  the  weights  now  found  for  top  and  bottom 
chords  and  girder  diagonals,  we  find 

Weight  of  girders  due  to  loads,  pounds, 


w 

h 


2  4-  0.97005^ 


0.970053/2  4- 

i.i64o63/2  4-  i-5o897/22 

1.358073/2  4-  i-7738i/22 

.1.552085/2  4-  2.03711^2 

1.746095/2  4-  2.29937/^2 

1.940105/2  4-  2.56093/22 

2.134116/2  4~  2.82196/22 

2.328126/2  4-  3.08260/22 

2.522137/2  4-  3-34295/22 

2.716148/2  4-  3-60304/22 

2.9ioi58/2  4-  3.86296^ 

3.104168/2    4-   4.I227I/22 

3.298179/2  4-  4-38235^ 
3.492190/2  4-  4.64i87/22 
3.686200/2  4-  4.90136^ 
3.88021 1/2  4-  5.16067^ 
4.O7422I/2  4-  5.41996/22 
4.268231/2  4-  5.67920/22 
4.462243/2  4-  5.93839^ 
4.656253/2  4-  6.19758/22 


0.916667/2  4-  2.69271^2 
1.138803/2  4-  4.33766/22 
I-35I563^2  +  6.27980^ 
1-558965^  +  8.51725/22 
1.763022/2  4-  11.05029/22 
1.964845/2  4-  13.87962/22 
2.165105/2  4-  I7-OO593/22 
2.364229/2  4-  20.42977/22 
2.562501/2  4-  24.15161/22 
2.760117/2  4-  28.17181/22 
2.957219/2  4-  32.49o63/22 
3.I53908/2  +  37.io829/22 
3.35O26i/2  4-  42.02499^ 
3-546340/2  4-  47.24089/22 
3.742190/2  4-  52.75607/22 
3.937844/2  4-  58.57070/22 
4-I33336/2  +  64.68470/22 
4.328684/2  -h  71.09829/^2 
4.523913/2  +  77.81148/22 
4.719046/2  4-  84.82431/22 
4-9i4o65/2  4-  92.I2937/22 


5 
6 

7 
8 

9 

10 
ii 

12 
13 
14 
15 

16 

17 

18 


20 

21 

22 

23 
24 


This  weight  of  girders  is  to  be  increased  by  one-tenth  of  itself, 
as  explained  in  article  165.  Also,  it  will  be  augmented  to  meet 
the  strain  brought  upon  the  top  chords  by  the  head  system. 

174.  Make  the  floor  of  2^-inch  oak  planks,  52  pounds  per 
cubic  foot,  and  of  the  width  of  q  feet.  Then,  if  q  =  16  feet, 

Weight  of  floor  =  —  X  52^  = /  pounds.     (573) 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


491 


175-  Longitudinal  I  Floor  Beams  ;  Conditions  and 
Weights  given  in  Article  157.  —  We  may  further  explain  the 
assumption  in  (501)  thus  :  Taking  an  analytical  table  of  ordi- 
nary wrought-iron  I-beams,  we  may  easily  see,  that,  for  depths 
of  8  inches  and  upwards,  we  have  approximately 

J  =  0.15,  (574) 

r  being  the  radius  of  gyration,  and  d  the  depth  of  beam.    Now, 
by  equation  (184), 


_   • 

S' 


2  I 


hence  if  we  take,  as  we  manifestly  may,  d  =  -  -,  and  eliminate 
r,  we  shall  find 


3  * 


=  —  S, 

ion 


as  in  (501),  where  the  same  notation  is  used,  d  being  in  inches, 
and  /  in  feet. 

We  obtain,  from  (503), 

Weight  of  6  wrought-iron  longitudinal  I-beams,  in  pounds, 


=  6Lt 


n 

O.I3OOOOO 

4 

O.IO4OOOO 

5 

0.0866667 

6 

0.0742857 

7 

O.O65OOOO 

8 

0.0577778 

9 

0.0520000 

10 

0.0472727 

n 

°-°433333 

12 

0.0400000 

13 

0.0371429 

H 

n   -  ji          v:>/:>/ 

n 

=  6Z/  +  /' 

0.0346667 

'5 

0.0325000 

16 

0.0305882 

17 

0.0288889 

18 

0.0273632 

19 

0.0260000 

20 

0.0247619 

21 

0.0236364 

22 

0.0226087 

23 

0.0216667 

24 

492 


MECHANICS  OF   THE   GIRDER. 


176.  Also,  let  the  transverse  I-beams  be  conditioned  as  in 
article  (158) ;  then  (507)  yields,  taking  5  from  (506), 


Weight  of  (n  —  i)  transverse  I-beams  due  load,  pounds, 

=  (n  —  i)  X  A  X  12  x   i8S 


(576) 


n 

5.2650  4-  I2 

0.003949  +  LI 

0.18225  +  <£ 

243 

4 

5.6160 

0.003369 

0.19440 

324 

5 

5.8500 

0.002925 

0.20250 

405 

6 

6.0171 

0.002579 

0.20828 

486 

7 

6.1425 

0.002303 

0.21262 

567 

8 

6.2400 

0.002080 

0.21600 

648 

9 

6.3180 

0.001895 

0.21870 

729 

10 

6.3817 

0.001740 

0.22091 

810 

ii 

6.435° 

0.001609 

0.22275 

891 

12 

6.4800 

0.001495 

0.22431 

972  j  13 

6.5186 

0.001397 

0.22564 

I053    X4 

6.5520 

0.001304 

0.22680 

H34 

15 

6.5812 

0.001234 

0.22781 

1215 

16 

6.6071 

o.  001166 

0.22871 

1296 

\I7 

6.6300 

0.001105 

0.22950 

1377 

18 

6.6505 

0.001050 

0.23021 

1458 

19 

6.6690 

O.OOIOOO 

0.23085 

1539 

20 

6.6857 

0.000955 

0.23143 

1620 

21 

6.7009 

0.000914 

0.23195 

1701 

22 

6.7148 

0.000876 

0.23244 

1782 

23 

6.7275 

0.000841 

0.23287 

1863 

24 

177.  Equation  (512)  becomes,  when  ;/  is  odd, 


Weight  of  iron  to  be  added  to  transverse  I-beams  on  account  of  wind, 
in  pounds, 


~  I2 


(577) 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


493 


25.000  -j — 

39.285 

53.703 
68.182 
82.692 


116.65 

n 
5 

256.62 

7 

451-01 
699-85 

9 
ii 

1003.11 

13 

i 
=  h 

97.222  - 

3 

1360.80 

n 

'5 

111.765 
126.315 
140.873 

1772.93 
2239.49 
2760.48 

21 

155435 

3335-91 

23 

—  —  - 


terms  in  summing  (509), 
for  the  two 


178.  When  n  is  odd,  we  use 

adding  4  X  -J(«a  —  i)  =  2f  n  — 

diagonals  of  the  middle  panel,  and  find,  as  in  (509), 

Weight  of  horizontal  diagonals,  in  pounds, 

n(n  —  i)  +  (n  —  i)(n  —  2)  } 

+  (n  -  2)(«  -  3) 

+  (n  -  3)O  -  4) 


=  4  X 


sin2  (/>, 


X  m  X 


n  —  i 


terms  +  J(«2  —  i) 

(578) 


=  h 

28.000  +  hi2 

0.0034568 

n 
5 

40.000 

0.0025195 

7 

51.852 

0.0019758 

9 

63.636 

O.OOI6232 

ii 

75.385 

0.0013767 

13 

87.111 

O.OOII949 

15 

98.823 

O.OOIO554 

17 

110.526 

O.OOO945O 

19 

122.222 

0.0008554 

21 

I33-9I3 

0.0007813 

23 

179.  In  summing  (515)  for  odd  values  of  * 

n  —  I 

1  IV  P  11°P 

terms  of  the  series,  and  add  \(n2  —  i)  for  middle  panel,  then 
multiply  the  sum  by  4,  since  the  two  wind  chords  are  to  be 
alike. 


494 


MECHANICS  OF   THE   GIRDER. 


Weight  of  wind  chords,  in  pounds, 


18 


n          2nq^Q 
=  0.006028164(2  +  -    —  —2 ^J 


n  -  i  +  2(n  -  2)  -f-  3(»  -  3) 

n  • 


2 


0.0150463 
0.0143414 
0.0138920 
0.0135871 
0.0133679 
0.0132030 
0.0130747 
0.0129721 
0.0128882 
0.0128183 


terms  + 


n  = 


(579) 


5 

7 

9 

ii 

13 
15 
17 
19 

21 

23 


180.  We  shall  now  have,  at  the  centre  of  each  vertical,  the 
load,  en,  as  defined  in  article  163.     Therefore 


Cross-section  of  lower  half  of  vertical  due  £«,  in  square  inches, 

£1       *O*C^      Cft  ^ 

Q3        1066 7 J 
Weight  of  all  lower  halves  of  vertical  due  £„,  in  pounds, 


=    2    X 


=    0.00021 


18        10667 
Cross-section  of  upper  half  of  vertical  due  en,  in  square  inches, 


12000 


(580) 


(581) 


(582) 


Weight  of  all  upper  halves  of  verticals  due  ent  in  pounds, 


=  2  x  -      X 


18        12000 


=  0.000185^2  —  -J/^«.      (583) 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          495 

As  in  the  second  part  of  article  163,  so  here,  suiting  the 
expression  to  the  changed  length  of  chord  segments,  we  have, 
from  the  assumed  wind  pressure,  the  moment 

62.5^  =  fS2  x  AsyBt,    B,  =  5667. 
Therefore 
Cross-section  of  any  vertical  due  bending-moment,  in  square  inches, 

(584) 


Weight  of  verticals  required  to  resist  bending-moment  due  wind,  in 
pounds, 

=  4  x  -^  X  £-  X  i2$y=  o.98o3922(  i  -  -\il.       (585) 
io        oon  \         n  I 

Adding  together  the  three  expressions,  (581),   (583),  and 
(585),  the  sum  is 

Weight  of  all  verticals,  pounds, 

=  /j  0.000393518^  -  iv  +  0.9803922^1  -  ~yi  (586) 


-h 


o.ooi  I52Z7  4-  1.5425^  4-  0.0000007 44/3  +  O.OOOO2495/2  -|-  0.956317  4-  0.0032 
o.ooi  I79Z7  4-  1.9654^  4-  o.oooooo699/3  -j-  o.oooo2oo2/2  -f  0.977477  -f-  0.0055 
o.ooi  194^7  +  2.3885^  +  o.oooooo677/3  -f  o.ooooi725/2  -j-  0.990437  -(-  o.ooSi 
o.ooi 203Z/  +  2.8077 L  -f  0.000000650/3  +  o.ooooi49o/2  +  0.998217  +  o.oi  1 5 
0.001209^7  -j-  3.2245^  +  0.0000006337 3  +  o.ooooi3io72  -f  1.003837  +  0.0152 
O.OOI2I3Z7  +  3.6396^  +  o.oooooo6i673  +  o.ooooi i6872  +  1.007707  -f-  0.0197 
o.ooi2i6Z7  -f  4.0536^  +  o.oooooo6o673  -\-  o.ooooio5472  -f-  i.oio6o7  -f  0.0245 
o.ooi  2i8Z7  -}-  4466SZ  +  o.oooooo59373  +  o.ooooo96o72  -f  1.012897  -f  0.0300 
o.coi22oZ7  -f  4.8793Z  -f  0.00000058 573  -f-  o.ooooo88i72  -f  1.014757  -f  0.0359 
o.oot22iZ7  +  5-29I4Z  -f  0.00000057773  +  o.ooooo8i472  +  1.016377  +  0.0425 
o.ooi 222Z7  4-  5-703iZ  +  o.oooooo57o73  +  0.000007 5672  +  1.017677  +  0.0494 
O.OOI223Z7  4-  6.II45Z  4-  o.oooooo56473  4-  o.ooooo7o672  4- 1.018857  4- 0.0571 
o.ooi 224Z7  4-  6-5257Z  4-  o.oooooo55973  4-  o.ooooo66372  4- 1.019927  4-  0.0651 
o.ooi 224Z7  4-  6-9367Z  4-  o.oooooo55573  4-  o.ooooo62372  4- 1.020907  4-  0.0739 


a 


9 


ro 


496  MECHANICS   OF   THE    GIRDER. 


=  h 


o.ooi225Z7  4-  7-3474Z  4-  o.oooooo549/3  +  o.ooooo59o72  4-  i. 021787  +  0.0829 
o.ooi225Z7  4-  7.5811^  4-  O.OOOOOO546/3  4-  o.ooooo553/2  4-  I.O2263/  +  0.0928 
O.OOI225Z/  4-  8.i686Z  -f  o.oooooo543/3  4-  o.ooooo53i/2  4-  1.023427  -j-  0.1029 
o.ooi226Z7  4-  8-5797Z  4-  o.oooooo539/3  4-  o.ooooo505/2  -f  i.o24i8/  -f-  o.i  138 
O.OOI226Z7  4-  8-9S94Z  4-  o.oooooo537/3  4-  0.0000048 3/2  4- 1.024917  -f  0.1250 
o.ooi226Z7  -f-  9-3997Z  +  o.oooooo533/3  4-  o.ooooo462/2  4-  1.025607  -f  0.1370 
o.ooi226Z7  4-  9.8o92Z  +  0.0000005327 3  +  o.ooooo44372  +  1.026277  4-  0.1492 


18 


20 


24 

since  for  the  even  values  of  n  we  have  en,  given  in  article  163, 
and  for  the  odd  values 

es  =  0.62430^7  +  1040.  5^  4-  o.oio82i72  4-  18.03537  +  o.ooi85o/;/24-  5.925^  4-  14.58-, 
e7  =z  0.44593Z7+  IO4O-5Z  4-  o.oo552i72  4-  12.88237  4-  o.ooi204>£724-6.i3i//  4-  21.38-, 
e9  =  0-34683Z7  +  1040.  5Z  4-  O.OO334O72  4-  10.01967  4-  0.000881  hi  2  4-  6.237^  4-  28.19-, 
ea  =  0.28378Z/  4-  1040.  5Z  4-  O.00223672  4-  8.19787  4-  o.ooo69i//724-  6.302^  4-  34.99-, 
£i3  =  0.2401  2Z7  4-  1040.  5Z  4-  o.ooi6oo72  4-  6.93667  4-  o.ooo  567/^7  2  4-  6.345/6  4-  41.80-, 
els  =  o.2o8ioZ74-  I040-5Z  4-O.OOI20272  4-  6.01  177  4-  0.000480/^7  2  4-  6.376^  4-48.60-, 
eJ7  —  o.i8362Z7  4-  1040.  5Z  +  o.ooo93572  4-  5.30457  4-  0.00041  67/7  2  4-  6.399/6  4-  55.40-, 
eX9  =  o.i6429Z74-  1040.  5Z  4-  o.ooo74272  4-  4.74617  4-  0.000366/^7  2  4-  6.417^  4-62.21-, 
e2i  =  O.I4864Z7  4-  1040.  5Z  4-  o.ooo6i372  4-  4.29407  4-  o.ooo327/z724-  6.432^  4-  69.01-, 

£23  =  0.1357  iZ7  4-  1040.  5Z  4-  0.00051  172  4-   3.92057  4-  o.ooo295///24-  6.444,4  4-  75.82-, 
=  the  load  applied  at  centre  of  each  support. 

Here,  as  in  article  163,  we  have  put  ^/for  h  in  the  last  three 
terms  of  the  value  of  tn  ;  a  substitution  introducing  no  practical 
error  in  the  small  resulting  terms,  but  enabling  us  to  keep  our 
final  equation  down  to  the  second  degree  with  respect  to  h. 

181.  For  the  head  lateral  system,  we  proceed  as  in  article 
164,  now  having  a  pair  of  diagonals  and  a  strut  for  each  panel 
so  far  as  the  head  system  extends,  say  to  n  —  6  of  the  central 
panels  when  head  room  is  sufficient. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          497- 

(;/  j\th 
-  J    panel  point  is, 


since  Wl  —  2,500-, 


2n    \  2/2  32        n 

=  78.125^  ~  T^/ (K  odd); 

(;Ath 
-J    panel  point, 

M  =  5 — •/(«  —  -  )-  =  — - 

2^     \  2/2  32 

=  78.125/z/  (w  even); 


=    78.125—  («  even); 

fi 

=  jy  x  -  =  156.25^^^  ^  (n  odd), 

;z  «3       ^x 

=  156.25 —  («even); 

nql 

requiring  each  diagonal  tie  to  resist 

I56-'5       X-  VL  x  ^iTL-1  pounds  (n  odd), 
cos « cos  <^r        ^x  «3 

or 

— T5  *2^ —  x  —  pounds  («  even), 
cos  a  cos  (/>!       «^x 

and  to  have  a  cross-section 

I56.25///(^  -    I) 

15000  cos  a  cos  (/>i/23^/ 

-  i)/z/ 
cos  0,      "   square  mches  ^l 


498 


MECHANICS  OF   THE    GIRDER. 


or 


O.OI04W 
S  =  nqt  cos  0,  square  mches  (*  even>;  (588) 


'calling,  as  before,  cos  a  =  i. 
iLength  of  each  head  diagonal  = 


n  cos  a  cos  (f>1        n  cos  <£, 


practically. 


Weight  of  2  (n  —  6)  wrought-iron  head  diagonals,  in  pounds, 


=  a(«  -  6)   X  jg  X  5^5-  X 


;z2  —  \)hl 


0.003858 


^  - 


o.oo3858O2  -  i)  (n  - 


cos  </>, 


(«  odd),      (589) 


0.00.385  8  (»  —  6)f  —  4-  324^)  («  even),    (590) 


n 

n 

-  +hl* 

4 

=  /; 

H.I999+/2/2 

0.00015363 

15 

- 

- 

5 

12.5000 

0.00015070 

16 

- 

0 

6 

13.70*3 

0.00014634 

17 

1.2245 

0.00007713 

7 

15.0000 

0.00014289 

18 

2.5000 

0.00012056 

8 

16.2049 

0.00013855 

19 

3-7037 

0.00014113 

9 

17.5000 

0.00013503 

20 

5.0000 

0.00015432 

10 

18.7074 

0.00013093 

21 

6.1983 

0.00015811 

ii 

20.0000 

0.00012754 

22 

7.5000 

0.00016075 

12 

21.2092 

0.00012375 

23 

8.6982 

0.00015885 

T3 

22.5000 

0.00012056 

24 

IO.OOOO 

0.00015747 

14 

>  a.  —  1  8  feet,  nnd    Z    —  T  4-  (  l%n\  —  T  i  tan2A 

CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


499 


Multiplying  the  cross-section,  (587),  (588),  of  head  diagonal 
by  2  X  1  0,000  cos  c^tan  <£„  we  find,  after  dividing  by  2,500 
pounds  inch  strain, 


o    = 


0.0104^  x   2  x   ioooo(«2  — 


i    ;r  —  i 


h  square  inches  (n  odd), 


12       n2 

0.0104!  x  2  x  ioooo///tan</>r 
2500;^ 

=  -f^h  square  inches  (n  even), 
=  cross-section  of  head  strut. 

Weight  of  (n  —  5)  head  struts,  in  pounds, 


(591) 


(592) 


5;    18            n2    12    V593 

„(/*  -  5)O2  -  i 

-h                (n  odd), 

n2 

=  &(*  -  5)^  X  12  x  1  8,             (594) 

=  5(n  —  5)^                      (n  even), 

n 

n 

- 

4 

=  h 

.49-7777 

15 

o 

5 

55.0000 

16 

5.0000 

6 

59-7924 

17 

9-7959 

7 

65.0000    1  8 

15.0000 

8 

69.8061 

19 

I9-753I 

9 

75.0000 

20 

25.0000 

10 

79.8186 

21 

29.7521 

ii 

85.0000 

22 

35.0000 

12 

89.8299 

23 

39-7633 

13 

95.0000 

24 

45.0000 

14 

500  MECHANICS  OF   THE   GIRDER. 

Calling  the  compression  along  each  segment  of  top  chord 
due  head  diagonals  equal  to 


l        or      156.25-. 
^ 


according  as  n  is  odd   or  even,  and  taking  the  allowed  inch 
strain  in  compression,  as  above,  viz., 

8000 

2     =  7529  pounds  =  3.764  tons, 

*"  40000 
we  have 

Cross-section  of  iron  to  be  added  to  segments  of  top  chord  in  head 
system,  square  inches, 

156.25     n2  -  i     hi 


=  S  =  0.00115295;^,  (595) 

156.25    hi 


=  0.00115295—.  (596) 

n 

Weight  of  added  iron  in  (n  —  6)  panels  for  top  chords,  pounds, 

=  2(«_6)  X  ^  X  ^S,  (597) 

18          n 

=  0.0076863—     6  ~  T^/«  (n  odd), 


(n  even), 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


501 


n 

n 

=  hi2 

- 

4 

=  hi2 

0.00030609 

15 

- 

5 

0.00030024 

16 

0 

6 

0.00029155 

J7 

• 

0.00015366 

7 

0.00028468 

18 

0.00024020 

8 

0.00027603 

19 

0.00028116 

9 

0.00026902 

20 

0.00030745 

10 

0.00026085 

21 

0.00031427 

ii 

0.00025409 

22 

0.00032026 

12 

0.00024654 

23 

0.00031648 

13 

0.00024020 

24 

0.00031373 

14 

182.  As  explained  in  article  165,  we  shall  here  augment,  by 
one-tenth  of  itself,  each  of  the  following  expressions  just  found; 
viz.,— 

The  girders  proper, 

The  vertical  supports,  and 

The  lateral  head  struts. 

Then,  adding  together  all  the  parts  of  the  complete  bridge, 
and  putting  the  sum  =  2,ooon  W>  the  weight  of  any  bridge  in 
pounds,  we  derive  the  following  values  of  W,  in  terms  of  L,  /, 
and  /i,  for  the  different  values  of  ;/. 

Then,  by  assigning  values  to  L  and  /,  differentiating,  and 

putting  =  o,  we  get   W  a  minimum^  and  h  best,  as  in 

dh 

article  140,  equations  (469)  and  (470). 


502 


MECHANICS  OF  THE   GIRDER. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


503 


1 
"T 


o 

"§    * 

O^         N 


n" 

5 


+ 


4 

vO 


DO 

tn 


504 


MECHANICS  OF   THE   GIRDER. 


I 


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+  + 


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d  + 


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+ 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


505 


CN  d 


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CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          507 


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MECHANICS  OF   THE   GIRDER. 


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CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


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512  MECHANICS  OF   THE   GIRDER. 

183.  Among  the  deductions  to  be  drawn  from  this  table,  for 
the  Brunei  double-bow  bridge  of  double  web  system,  are  the 
following  :  — 

1st,  For  a  given  uniform  live  load, 

n  oc  /*  nearly;  (620) 

and  generally 

n  oc  (JJ     x  /*  nearly.  (621) 


2d,  For  spans  less  than  400  feet, 


For  spans  of  400  feet  and  upwards, 

nearly-  (623) 

3d,  For  different  spans  with  same  live  load  per  running-foot, 

W  oc  Ih  nearly;  (624) 

and  for  the  same  span  under  different  uniform  live  loads, 

nW  oc     ~       nearly.  (625) 


Many  other  conclusions  may  be  drawn  from  this  table,  and 
weights  of  intermediate  spans  may  be  derived  by  interpolation  ; 
but  the  equations  (598)  to  (619),  inclusive,  cover  the  whole 
case. 

184.  EXAMPLE.  —  We  now  proceed  to  find  the  strain  sheet 
for  the  2oo-feet  span  of  the  table  in  article  182  in  a  manner 
similar  to  that  employed  in  article  169. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          513 

We  now  have 

/  =  200         feet,  q  =  1  6  feet  ; 

//  =  40.788  feet,  gl  =  18  feet. 

n  =  13. 

tiL  =  200         tons,  L  =  15  A  tons  > 

n\V  =  103.288  tons,  W  =  7.945  tons; 

/r  -h  Z  =  23.330  tons. 


Weight  of  floor,  by  article  174,  ^  X  200  =     34667  pounds  ; 
Total  live  load,  //Z,  =  400000  pounds. 

Total  load  on  longitudinal  I-beams  =  434667  pounds. 

Load  on  each  panel  length  of  every  longitudinal  I-beam  spaced  3.2  feet 


_         x  =  6687  pounds. 

Then,  by  (502), 

Cross-section  of  beam 

434667 
=  S  =  0.00015  x    -  =  5.0154  square  inches. 

Take  b  =  4  inches  =  breadth  of  flange. 
b  —  b,  =  0.26  inch  =  thickness  of  web. 
Then,  from  (552),  (551),  and  (550), 

d  —    9-774  inches  =  depth  of  beam, 
//  —  tft  =    0.66  1  inch      =  depth  of  two  flanges, 

/=  75.416  =  moment  of  inertia  of  section, 

which  is  lur-vr  than  /  for  the  sections  given  by  ordinary  beams 
of  the  same  area  of  section. 


ht  of  longitudinal  I-beams,  6  in  number, 

=  6  X  -j^  X    12   X    200  X  5.01538  =  20062  pounds. 


5  14  MECHANICS  OF   THE   GIRDER. 

Upon  the  transverse  I-beams  we  have 

Live  load,  400000  pounds, 

Floor,  34667  pounds, 

Longitudinal  I-beams,    20062  pounds. 

Total  for  13  panels,       454729  pounds. 
Load  on  i  beam,  34979  pounds. 

From  (504), 

/       12  x  18  X  34979 

d  =    8  x  2  x  IQOQO    =  47.2216  =  2S, 

by  (505)  ; 

.*.     S  =  23.6108  square  inches  for  vertical  load, 
and  for  the  wind  pressure, 

40.788 
Wi  =  2500  x  -  =  7844  pounds  per  panel, 

=  7542'5 


=  7.6798  square  inches,  ist  and  i2th  beams; 
=  6.4532  square  inches,  2d  and  nth  beams; 
=  5.3332  square  inches,  3d  and  loth  beams  ; 
=  4.3199  square  inches,  4th  and  gth  beams; 
=  3.4132  square  inches,  5th  and  8th  beams; 
=  2.6133  square  inches,  6th  and  7th  beams. 

The  total  cross-sections  for  each  half-span  are 

S  =  31.2906  square  inches, 
=  30.0640  square  inches,* 
=  28.9440  square  inches, 
=  27.9307  square  inches, 
=  27.0240  square  inches, 
=  26.2241  square  inches. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          515 

Satisfying  the  condition  (505),  we  may  assign  values  to  d 
and  dv  and  use  (557)  and  (558)  in  finding  the  thickness  of  each 
transverse  beam. 

Put  2  light  12-inch  beams  at  each  panel  point,  the  section 
of  each  being  \S.  Then  we  have 

d  =  12  inches  =  depth  of  beam, 
d  —  di  =     2  inches  =  depth  of  2  flanges ; 

and  (558)  becomes,  for  breadth  of  flanges, 

b  =  (24  ~  I2  -f  — )  X  iS  =  0.17803^  =  5.5707  inches, 

y  2    X    2  2  1 2  J 

=  5-3523  inches, 
=  5.1529  inches, 
=  4.9725  inches, 
=  4.8111  inches, 
=  4.6687  inches; 
and  (557)  gives 

bl  =  iQ4x  2"  2"*  X  %S  =  °'l6363^  =  5-I2°3  inches, 

=  4.9083  inches, 
=  4-7363  inches, 
=  4.5705  inches, 
=  4.4221  inches, 
=  4.2912  inches. 
Thickness  of  web  =  0.4504  inch  =  b  —  bly 

=  0.4440  inch, 

=  0.4166  inch, 

=  0.4020  inch, 

=  0.3890  inch, 

=  0.3775  inch. 

The  weight  of  these  24  transverse  I-bearhs  is 
12  x  1 8  x  T5s^  =  205  7 7  pounds^ 

since  25  (—  sum  of  all  the  cross-sections)  is  342.9548  square 
inches. 


516  MECHAATICS  OF   THE    GIRDER. 

Cross-sections  of  horizontal  diagonals  are  found  by  dividing 
the  strains  in  (508)  by  15,000,  where  we  now  have 

W,  =  7844,     —  -^-  =  0.52293, 
15000 

sin^j  =  0.76017, 

0.^2203 
5  =  ,  X   13  X  o.76o,7('3  X   »,  i»  X   i,, 

ii  X  10,  10  x  9,  9  X  8,  8  x  7,  7  X  6) 
=  0.026458  X  156  =  4.1274  square  inches, 
=  0.026458  X  132  =  3.4925  square  inches, 
=  0.026458  X  no  =  2.9104  square  inches, 
=  0.026458  X  90  =  2.3812  square  inches, 
=  0.026458  X  72  =  1.9050  square  inches, 
=  0.026458  X  56  =  1.4816  square  inches, 
=  0.026458  x  42  =  i.i 1 12  square  inches, 

for  the  respective  panels. 

.*.     2S  =  67.4150  square  inches, 
and  weight  of  26  horizontal  diagonals  is 

I2.X.1     X   5-  X  67.415  =  5321  pounds. 
sin<£,         1 8 

The  cross-section  of  each  panel  length  of  each  wind  chord 
is  given  by  (514),  thus, 

7844    X    200 


2   X   13   X   18  X  6400V" 
3  X 
=  °-52377  X  12  = 
=  °-52377  X  22  = 
=  o-52377  X  30  = 
=  0-52377  X  36  = 
=  °-52377  X  40  = 
=  °-52377  X  42  = 
=  o-52377  X  42  = 

^,  ^  /\  a.  i, 

10,  4  x  9,  5  X  8,  6 
6.2852  square  inches, 
11.5229  square  inches, 
15.7131  square  inches, 
18.8557  square  inches, 
20.9508  square  inches, 
21.9983  square  inches, 
21.9983  square  inches. 

=  106.3253  square  inches. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES, 


These  sections  can  easily  be  made  up  of  channels  and 
plates,  or  of  beams  and  plates,  with  the  required  radius  of 
gyration  given  in  article  162. 

In  summing  these  sections  for  the  weight  formula,  all  are  to 
be  taken  four  times,  except  the  last,  which  is  taken  twice  only. 

12  x  200  x  106.3253       5 
Weight  of  wind  chords  = X  yg  =  21810  pounds. 

O 

Supported  by  verticals,  we  have 

Live  load,  400000  pounds, 

Floor,  ^34667  pounds, 

Longitudinal  I-beams,  20062  pounds, 
Horizontal  diagonals,  5321  pounds, 
Wind  chords,  21810  pounds. 

481860  —  26  =  18533 

Transverse  I-beams,        20577  -r-  24  =       857 

Weight  on  each  vertical  =  zn  =  19390 

Therefore  we  have  the  cross-sections,  by  (580), 

=  1.81776  square  inches; 

±w \j  i 

by  (582), 

I039O 

Sl  =  -       -   =  1.61583  square  inches; 
12000 

for  the  lower  and  upper  halves  respectively  of  the  verticals  due 
load;  and,  for  the  bending-moment  due  wind,  (584)  gives 

S2  =  —5 ^^  =  ^-3Q'?67  square  inches, 

68  x   13 

Section  of  compressed  half  =  5.21143  square  inches, 
Section  of  extended      half  =  5.00950  square  inches. 

And,    since   the    upper   and   lower   halves   of   the   girder   are 
symmetrical,    and    the    sum    of    the   lengths    of    the   verticals 

=  ^y  =  %h(n  —  - )  by  (521),  =  J  X  40.788  X  ^g8-,  we  have 


518  MECHANICS  OF   THE   GIRDER. 


=  2  X  -5-   x  5.21143   X  —  X  40.788  X  -  —  =     6103  pounds, 
lS  3  J3 


Weight  of  lower  halves 

=  2  X  -5-   x  5. 
lS 

Weight  of  upper  halves 

=  2  X  -^-   X  5.0095      X  —  X  40.788  X  -  —  =     5866  pounds. 
18  3  13         _ 

Total  weight  =  11969  x   — 

10 

=  13165  pounds. 
after  adding  -fa  for  braces,  etc. 

The  sections  may  be  made  up  of  2  channels,  the  one  ver- 
tical, the  other  inclined  at  an  angle  whose  tangent  is  ^. 

According  to  the  principles  of  article  165,  the  bars  in  the 
bracing  of  these  supports  should  have  a  cross-section  of  about 
-|  inch  ;  that  is,  about  J^  of  (5  +  S2)> 

Equation  (587)  gives  the  cross-section  of  each  head  diagonal 
thus  : 


S  =  ____j^__     = 

6  X  i33  X   18  X  0.64972 

From  (589)  comes  the  weight  of  14  head  diagonals  in  the 
seven  central  panels  equal  to 

14  x  -**-  X  —   X  —  2°°      X  0.5556  =  614  pounds. 
18       13       0.64972 

Cross-section  of  head  struts,  by  (591), 

=  I68  X  4^788  =          8  incheS) 

12     X     169 

requiring  2  light  4-inch  channels  latticed  not  less  than  4  inches 
apart. 

Weight  of  8  head  struts, 

-xSx-5-Xi2Xi8x  3.3789  =  1784  pounds, 
10  18 

after  adding  T^  for  bracing. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES.          519 

Increment  of  section  of  top  chord  due  to  head  diagonal 
strain  is  given  by  (595),  thus  : 

156.25  x  168  x  40.788  x  200 
S  =  7529  X   i33  X  18  =  °'7I92  Square  mch> 

the  strain  being  =  0.7192  X  $$$£  —  2.707  tons. 
Weight  added  to  top  chords 

=  2  x  7  X  -£-  X  T2  X  2°°  X  0.7192  =  516  pounds. 
18  13 

For  each  of  two  girders,  the  horizontal  component  of  chord 
strain  is,  by  (564), 


and  the  chord  strains  are 

P  =  -    —  =  U  =  -    —JT  =  99.317  tons,  ist  panel; 

=  97.415  tons,  2d  panel; 
=  95.830  tons,  3d  panel; 
=  94.580  tons,  4th  panel : 
=  93.672  tons,  5th  panel; 
=  93.130  tons,  6th  panel; 
=  92.947  tons,  7th  panel. 
Cross-section  of  top  chord  due  load 

n 

=  26.381  square  inches,  ist  panel; 


=  25.876  square  inches,  2d   panel ; 

=  25.455  square  inches,  3d   panel; 

=  25.123  square  inches,  4th  panel; 

=  24.882  square  inches,  5th  panel ; 

=  24.738  square  inches,  6th  panel ; 

=  24.689  square  inches,  7th  panel. 

Augment  4th,  5th,  6th,  7th,  8th,  9th,  loth,  by  0.7192. 


520 


MECHANICS  OF   THE    GIRDER. 


Cross-section  of  bottom  chord  =  —  =  19.863  square  inches,  ist  panel; 

=  19.483  square  inches,  2d  panel ; 
=  19.166  square  inches,  3d  panel ; 
=  18.916  square  inches,  4th  panel ; 
'=  18.734  square  inches,  5th  panel ; 
=  18.626  square  inches,  6th  panel ; 
=  18.589  square  inches,  yth  panel. 

The  top  chord  may  be  composed  of  2  Q-inch  channels  and  i 
plate ;  the  bottom  chord,  of  3  bars  and  2  bars  in  alternate 
panels. 

From  (561),  we  find 

2sec2«  =  27.4329; 
and  (566)  gives,  adding  -J^, 


Weight  of  top  chords  =  —  x 
10 


2002 
40.788 


X  27.4329 


=  38206  pounds. 


From  (567),. 

Weight  of  bottom  chords  =  — X  38206  =  28767  pounds. 

From  (571),  the  strain  on  a  girder  diagonal  is  called 
'  T5r\  X  200  x  1.8  x  ^0  =  2<I2l66sec^ 

Section  =  |Z  =  1.16  square  inches. 

=  1.52  square  inches. 

=  1.83  square  inches. 

=  2.06  square  inches. 

=  2. 20  square  inches. 

=  2.24  square  inches. 

=  2. 20  square  inches. 

=  2.06  square  inches. 

=  1.83  square  inches. 

=  1.52  square  inches. 

=  1. 1 6  square  inches. 


2   x  8  X  40.788 

=  3.1023  tons,     2d   panel. 

=  4.0600  tons,    3d   panel. 

=  4.8790  tons,    4th  panel. 

=  5.4861  tons,    5th  panel. 

=  5.8564  tons,    6th  panel. 

=  5.9807  tons,     7th  panel. 

=  5.8564  tons,    8th  panel. 

=  5.4861  tons,    9th  panel. 

=  4.8790  tons,  loth  panel. 

=  4.0603  tons,  nth  panel. 

=  3.1023  tons,  1 2th  panel. 


CALCULATION  OF   THE    WEIGHT  OF  BRIDGES. 


521 


These  sections,  being  in  alternate  tension  and  compression, 
may  be  made  up  of  4  angle  irons,  \\  X  i£,  latticed  at  such  a 
distance  apart  that  the  unsupported  length  may  be  not  more 
than  one  hundred  times  the  radius  of  gyration  of  the  section. 

The  weight  of  these  girder  diagonals  is,  from  (572),  equal  to 


4  X  12  x  5  X  0.45  X  2003  x  3 
8  X  is2  X  18  X  40.788  X  8 


ii  =      6      ssec'0  x  U 
10  10 

=  21088  pounds, 


since  2sec2<9  =  58.7324  by  (563),  and  we  increase  by  one-tenth 


for  latticing  and  attachments. 


STRAINS  AND  CROSS-  SECTIONS. 

TONS.  SQUARE  INCHES. 


84.555- 


83.672 


83.  J30       92.947 


FIG.  119. 


For  each  of  two  girders.  Span,  200  feet.  Central  height,  40.788  feet.  Uniform  live  load,  i  ton  =  2,000 
pounds  per  linear  foot,  applied  at  centres  of  verticals;  the  verticals  in  this  case  acting  merely  as 
struts  in  the  lower  half,  and  as  suspenders  in  the  upper  half.  The  diagonals,  only  one-half  of 
them  being  shown  in  the  figure,  are  alternately  in  tension  and  compression;  the  greatest  strains 
being  the  same  on  each  of  the  two  diagonals  of  a  panel.  Bridge  weight  =  103.288  tons. 

The  deflection  due  to  full  load  is  found  for  any  point  by 
equation  (559),  having  now 

Bl  =  4.380  tons  per  square  inch, 

E  =  12000         tons  per  square  inch, 
h,  =  h    =         40.788  feet, 
a  =  \l  —       100         feet, 

and  x  being  measured  from  centre  of  span  ; 


522  MECHANICS  OF   THE   GIRDER. 

.'.     Deflection  D*  =  1.490  inches  for  x  =        o,  centre; 

D2  —  1.483  inches  for  x  =     100  -f-  13  ; 

Dz  =  1.432  inches  for  x  =     300  -^  13  ; 

D4  =  1.327  inches  for  x  =     500  -f-  13  ; 

JDS  =  1.162  inches  for  x  =     700  -^  13  ; 

/?6  =  0.815  inch     f°r  x  =    9°°  ~^  X3  > 

/?7  =  0.583  inch     for  #  =  noo  -f-  13  ; 

Z>8  =  o  inch  for  #  =  1300  -s-  13,  ends. 

Equation  (366)  yields  the  excess  of  length  required  in  top 
chord  to  give  the  proper  camber, 


x  2°5'34  x  I2  = 


since  length  of  polygonal  top  chord  is  equal  to 


—  X  2seca  =  —  X  13.3466  =  205.34  feet; 
13  !3 

.-.     Mean  excess  per  panel  =  --  =  0.138  inch, 
or  a  little  more  than  4  inch. 


BRIDGES   OF  CLASS  II.  $2$ 


CHAPTER    XI. 

BRIDGES  OF  CLASS  II.  —  BEST  NUMBER  OF  PANELS  AND  BEST 
HEIGHT  DETERMINED  FOR  A  GIVEN  SPAN  UNDER  A  GIVEN 
UNIFORM  LIVE  LOAD.  —  LEAST  BRIDGE  WEIGHT  AND  LIMITING 
SPAN  FOUND. 

SECTION  i. 


The  Parabolic  Bowstring  Girder  of  Double  Triangular  System  (Fig. 

with  the  Extreme  Diagonals  omitted,  and  a  Vertical  Suspender  at 
Extreme  Panel  Point. 

185.  Let  /  =  span,  in  feet. 

//  =  height  of  girder  at  centre,  in  feet. 
;/  =  number  of  panels. 

'  L  :=  panel  weight  of  uniform  live  load,  in  tons. 
W  =  panel  weight  of  bridge,  in  tons. 

The  height  of  girder  at  any  point,  ;r,  is  given  by  (472),  and 
at  all  vertices  by  (473),  if  we  make  r  =  i  for  the  first  point, 
and  put  2/1  for  h  throughout,  thus  : 

(626) 

+  i)(«  -r-  i), 
-  >  =  ^(n  -  2r  -  i),  (627) 

tan«    =  by  -r-  -  =  ^(n  -  2r  -  i),  (628) 

n       nl 

sec2«  =  i  +  tan2**  =  i  +  ^-(n  —  2r  —  i)2,  (629) 

n2l2 


524 


MECHANICS  OF   THE    GIRDER. 


tan#r  =  —tan 


.,  =  -y,  «-  '-  =  -4-(«  _  r), 


i6/r 


?0  =  i  +  tan26>  =  i  4-  -y^O  ~  ?*)2- 


(630) 
(631) 


I  a,  #,  and  0  are  defined  in  article  49.  •  , 

The  live  load  and  a  large  part  of  the  dead  load  are  applied 
at  the  panel  points  of  the  bottom  chord,  and  are  transmitted 
by  the  diagonals  to  the  parabolic  arch,  which  is  equilibrated  by 
the  uniform  load,  leaving  only  a  tensile  strain  on  the  diagonals 
from  full  uniform  load.  We  shall  assume  that  the  two  diag- 
onals which  support  any  panel  weight  of  uniform  dead  load 
carry  each  one-half  of  the  same. 

186.  Moments  at  all  panel  points  due  ;/(  W  +  L)t  the  total 
load,  are,  from  equation  (65), 

W  + 


211 

and  the  horizontal  component  of  chord  strain  under  same  load 
is  equal  to 

If  =  M  =  i(^4-Z)^, 

as  in  (564),  and  is  uniform  throughout  for  maximum. 

H    } 


Greatest  strains  in  top  chord  =  P 

cos  « 

Greatest  strains  in  bottom  chord      =  U  —  H 

Cross-section  of  top  chord,  ,S          =  P  -5-  Q 
Cross-section  of  bottom  chord,  ,5,  =  U  -f-  T 


\ 


Take 


=  3.7647  tons 


(632) 


(633) 


(634) 


1   40000 
T  =  5  tons 

as  the  allowed  inch  strains  on  top  and  bottom  chords  respec- 
tively. 


BRIDGES   OF  CLASS  IL 


525 


I 

Length  of  segment  of  top  chord  =  —     -  inches, 

72  COS  « 


Volume  of  segment  of  top  chord 


n  Q  cos2  « 


-  =  \(W  +  L) 


Q/i  cos2 « 


cubic  inches, 


(635) 


by  summing  (629)  for  values  of  r  from  o  to  ;/  —  I,  inclusive. 

Therefore,  calling  weight  of  a  cubic  inch  of  wrought-iron,  as 
in  all  cases,  -f$  pound,  we  find 


Weight  of  top  chords,  in  pounds, 
=  £  X 


(636) 


12  x   3-7647/M 

3/2  V       n)  \ 

r.^MH- 

71/2/2  4-  0.59028^  n 

n 

W  4-  Z 

0.885417/2  4- 

4.64844/^2 

8 

k 

0.996095/2  4- 

5.24688/^2 

9 

1.106772/2  4- 

5.84372^ 

10 

1.217449/2  4- 

6.43936^ 

ii 

1.328126/2  4- 

7.03412/^2 

12 

1.438804/2  4- 

7.62820/^2 

T3 

1.549481/2  4- 

8.22i68/z2 

14 

1.660158/2  + 

8.81480^ 

15 

1.770835/2  4- 

9.40752/^2 

16 

1.881512/2  -f 

IO.OOOOO//2 

i7 

1.992190/2  4- 

10.59216/^2 

18 

2.102867/2  4- 

11.18440/^2 

19 

2.213544/2  4- 

11.77600/^2 

20 

526 


MECHANICS   OF   THE   GIRDER. 


Weight  of  bottom  chords,  in  pounds,  =  -^-  X  — — ^—  X  nl2,    (637) 

1 8        8       Tli 

W  4-  Z       l*n 


w  +  z 


n 

0.666667/2 

8 

O.75OOOO/2 

9 

Q.833333/2 

10 

o.9i6667/2 

ii 

.oooooo/2 

12 

•OS3333/2 

I3 

.i66667/2 

14 

.25OOOO/2 

15 

•333333/2 

16 

•4i6667/2 

17 

.^ooooo/2 

18 

.66666  7/2 

20 

187.  The  Girder  Diagonals.  —  Separating  the  double  sys- 
tem into  the  single  web  systems  of  Fig.  35*2  and  Fig.  27,  let  us 
consider  first  that  of  Fig.  350,  and  find  the  difference,  A//,  of 
horizontal  strains  at  the  foremost  end  of  the  advancing  uniform 
discontinuous  load,  nL,  and  for  the  same  instant  at  the  next 
two  forward  panel  points  of  the  double  system. 

Putting  Z  for  W  in  equation  (60),  and  taking  r2  =  — J,  we 
find  the  moment  at  any  point,  x,  at  or  before  the  foremost  end, 
to  be 

Mx  =  —  (r  +  £)2(/  -  x),  (638) 

2/ 

where  c  =  length  of  whole  interval  in  the  single  system  =  — 

n 

in  the  double  system,  r  +  \  —  number  of  panel  points  of  bot- 
tom chord  loaded,  x  =  distance  from  left  end  to  the  point 
where  moment  is  taken. 


BRIDGES  OF  CLASS  II. 


527 


From  (472),  putting  2/1  for  //, 


(639) 


Therefore  the  simultaneous  horizontal  strains  due  live  load 
at  the  three  consecutive  panel  points,  of  which  the  foremost 
end  of  live  load  is  at  the  rear  one,  are 

H  =  ^  =  —     (r  +  i)2 

'    y    "  8/1  '         x 

£1    (f     _|_      IV 

=  -  -*£-  if  x  —  re,  foremost  end ; 

8/1         r 


(640) 


==  ^ 
8/1 

_Ll  (r 


(r  4-  -|)     if  x  =  (r  +  %)c,  next  panel  point; 
if  x  =  (r  4-  i)^,  next  panel  point; 


I 


where  r  takes  the  successive  values  J,  f,  f  ,£,... 

Then  we  find  the  greatest  differences,  that  is,  the  greatest 
horizontal  component  of  diagonal  strain  due  live  load,  thus  : 


1  =  _//  r  +  £ 
I  16/2  r  +  i 


(641) 


for  the  first  single  system  ;  r  taking  the  successive  values  J,  f , 
f,  J,  etc. 

Similarly,  for  the  second  single  system,  we  get  greatest 
horizontal  component  of  diagonal  strain  due  live  load, 


i6/i  r  4- 
Ll       r 


16/1  r  + 


,    (642) 


where  r  becomes  r,  2,  3,  4,  etc. 


528  MECHANICS   OF   THE    GIRDER. 

Now,  if  we  consider  the  horizontal  components  of  the 
two  diagonals  in  the  same  panel  of  the  double  system,  Fig. 
35,  with  its  extreme  tie  made  vertical,  we  see  that  ^H  of 
(641)  and  A|//  of  (642)  belong  to  the  odd  panels,  and  ^H 
of  (641)  and  A!//"  of  (642)  belong  to  the  even  panels.  Also, 
for  the  odd  panels,  r  of  (641)  is  less  by  ^  than  r  of  (642)  ;  and, 
for  the  even  panels,  r  of  (641)  is  greater  by  J  than  r  of  (642). 

Therefore,  reducing  so  that  r  belongs  to  the  second  system, 
we  find 

LI       r 
Compression,  AxZf  = 


odd  panels,      (643) 
Tension, 


Tension,  T 


1  67*  r  +  -I  - 

j.  even  panels,     (044) 

LI       T 

Compression,  ^H  =  -- 

i6h  r  -f  \  } 

which  expressions  are  identical  ;  and  the  sum  of  either  pair  is 


as  already  given  by  (570)  for  the  total  horizontal  component 
of  diagonal  strain  due  live  load  in  any  panel. 

Since  these  diagonals  are  to  be  alternately  in  tension  and 
compression,  the  load  travelling  either  way,  and  our  specifica- 
tions would  multiply  the  compressive  strains  by  1.8,  we  shall, 
for  convenience,  take 


due  live  load  for  each  diagonal  in  a  panel,  instead  of  multiply- 


BRIDGES  OF  CLASS  If,  529 

ing  by  1.8,  and  shall  treat  all  diagonals  as  in  compression  under 
the  inch  strain, 

Q  =  4 =  ?  tons. 

3 


2OOOO 

This  procedure  varies  a  little  from  the  specifications,  but  on 
the  safe  side,  since 

27*  -f-  i  >  i.Sr. 

Moreover,  since  the  dead  load,  with  the  exception  of  the  top 
chords  and  head  system  and  wind  braces,  is  suspended  at  all 
times  on  two  diagonals  which  transmit  it  to  the  equilibrated  top 
chord,  and  since  the  tensile  section  will,  for  practical  spans, 
not  be  greater  than  the  compressive  section  due  to  live  load 
as  above  augmented,  and  to  be  provided  for  in  compression,  we 
may,  as  appears  below,  leave  the  tensile  strain  which  will  come 
upon  the  diagonals  acting  as  suspenders,  almost  entirely  to  the 
material  put  into  them  to  resist  maximum  compression. 

It  is  to  be  observed,  that,  when  the  live  load  is  fully  on  the 
bridge,  there  is  no  compression  on  the  girder  diagonals,  but 
each  one  acts  simply  as  a  suspender  to  transmit  ^(  W2  +  L)  to 
the  equilibrated  top  chord  ;  W2  being  that  part  of  the  dead  load 
at  any  lower  apex.  Now,  from  the  results  tabulated  in  Chap.  X., 
we  may  doubtless,  in  the  present  case,  for  spans  not  over  600 
feet,  consider  W2  as  ranging  from  |  W  to  f  W,  while  W  ranges 
from  ^L  to  2L  approximately.  Therefore  \(  W2  +  L)  ranges 
from  \%L  to  \L  nearly,  in  spans  from  100  feet  to  600  feet. 

Taking  the  greater,  |Z,  as  the  vertical  component  of  ten- 
sion on  each  girder  diagonal,  we  have 

U   .  n 


SZcot0  =  ^-  X  ^-  ^ 

24         h         r(n  —  r) 


as  the  horizontal  component  of  tension  on  diagonals  acting  as 


530  MECHANICS  OF   THE   GIRDER. 

suspenders  ;   the  greatest  value  of  which  is  found  when  r  =  i, 
or  r  =  n  —  i.     That  is, 

x  =  L&.  _JL_  for  tension. 
24      /* 


A./7  = for  compression. 

Dividing  A//"max  and  A//"  by  5  and  by  f,  the  allowed  inch 
strains  in  tons  respectively  for  tension  and  compression,  and 
multiplying  by  sec  0,  we  find  these  resulting  cross-sections  for 
comparison : 

•S1!  =  —  X  .  X  — - —  nearly 

17       hcQsO       n  -  i 

.    (646) 


64       h  cos  d 

Now,  Sr  will  be  greater  than  5  only  when  r  =  i,  or  r  =  n  —  i ; 
that  is,  the  girder  diagonals  which  meet  at  the  first  and  second 
and  at  the  (n  —  2)°*  and  (n  —  i)th  lower  apices,  will  need  addi- 
tional section  under  full  load  to  the  extent  of  the  difference 
between  5r  and  5.  And  this  we  shall  supply  in  the  vertical 
braces  at  these  points. 


^ 

Cross-section  of  a  girder  diagonal  =  S  =  —  «  --  -;     (647) 

64/7  cos  v 

(648) 


Weight  of  2(n  —  2)  girder  diagonals,  in  pounds, 

=  2X—  x-S-X-^-X  - 
n          18        64        h 


i6nh\  #2/2\3O  30 

_  ZjsQi-  ay  ,    (if       5 

h\        i6n 


BRIDGES   OF  CLASS  II. 


531 


z 

=  — 

h 

o-234375/2  + 

10.  1  8  ^  ^  5^2 

8 

0.243056/2  + 

1  3  .05899'^ 

9 

0.250000/2  + 

l6.26OQO/22 

10 

0.255682/2  + 

19.  78964^ 

ii 

0.260417/2  + 

23.64873^ 

12 

0.264423/2  + 

27.83796** 

13 

0.267857/2  + 

32.35788/22 

14 

o.27o833/2  + 

37.20889/22 

15 

0.273437/2  + 

42.39I36/22 

16 

Q.275735/2  + 

47-90556/22 

17 

o.277778/2  + 

53-75I72^2 

18 

o.2796o5/2  + 

59.93002/22 

19 

0.28I250/2    + 

66.44O62/22 

20 

Adding  together  (636),  (637), 

and  (648),  we  find 

Weight  of  girders  due  to  loads,  pounds, 

_  Z 

=  h 

i.786459/24-  M-83399/22 

+ 

-h 

,55,084^ 

4.64844^ 

ft 
8 

1.989151/2  +  18.30587^ 

I.746095/2  + 

5.24688/22 

9 

2.I9OIO5/24-  22.IO372/22 

I.94OIO5/2  + 

5.84372^ 

10 

2.  389798^  4-  26.22900/22 

2.i34ii6/2  + 

6.43936^2 

ii 

' 

2.58S543/2  4-  30.68285^ 

2-328I26/2  + 

7.O34i2^2 

12 

2.  786560^  +  35.46616/22 

2.522I37/2  + 

7.62820^ 

13 

2.984005/2  4-  4O-57956/22 

2.7i6i48/2  + 

8.22168/1* 

14 

3.i8o99i/2  4-  46.  02369/1* 

2.9IOI58/2  + 

8.81480^ 

15- 

3.377605/2  4-  51.79888/^2 

3.i04i68/2  + 

9.40752/^2 

16 

3-5739I4/2  +  57.90556^ 

3.298179/2  +  IO.OOOOO/22 

I7 

3.769968/2  +  64.34388^ 

3.492190/2  +  I 

0.59216^ 

18 

3.965805/2+  71.11442/^2 

3.686200/2  +  11.18440^ 

19 

4-i6i46i/2+  78.2i662/22 

3.88021  1/2  +  11.77600^ 

20 

To  be  augmented  by  one-tenth. 


532  MECHANICS   OF   THE    GIRDER. 

188.  Floor  to  be  the  same  as  in  article  174. 

Weight  of  floor  =  &p/  pounds.  (649) 

189.  Take  longitudinal  I  floor  beams,  as  in  articles  157,  175, 
equation  (575). 

190.  The  transverse  I-beams  supporting  live  load,  floor,  and 
longitudinal  beams  are  here  conditioned  as  in  article  158,  and 
their  cross-section  due  vertical  forces  is  given  by  equation  (506). 
Their  weight  due  same  forces  is  given  by  (576). 

The  cross-sections  of  transverse  I-beams  due  to  wind  are 
expressed  in  (511). 

The  weight  of  iron  to  be  added  to  the  transverse  I-beams 
on  account  of  wind,  is  given  by  (512)  and  (577)  for  n  even  and 
n  odd. 

The  whole  effect  of  wind  pressure  is  to  be  transferred  to 
the  horizontal  system,  in  the  plane  of  the  bottom  chords,  by 
means  of  vertical  braces  connecting  each  transverse  beam  with 
both  top  chords. 

It  may  be  observed,  that  in  this  and  all  like  cases  a  shear- 
ing-stress is  generated  throughout  the  transverse  beam  by  the 
wind  pressure  transmitted  through  these  vertical  braces ;  but 
there  will  be  sufficient  reserve  material  in  the  web  of  the  beam 
to  resist  this  shearing-stress,  as  becomes  evident  on  reflection. 

191.  If  we  divide  equation   (508)   by   15,000,  we  have  the 
cross-section   of  any  horizontal   diagonal   in  the   floor  system. 
And  equations  (509)  and  (578)  give  us  the  weight  of  the  hori- 
zontal diagonals,  in  pounds,  for  the  even  and  odd  values  of  n. 

192.  For  wind  chords,  let  us  use  the  bottom  chords  of  the 
girders,   augmenting   their   cross-sections    by  the    quantity  in 
(513)  divided  by  the  tensile  inch  strain,  10,000  pounds. 

Although  this  augment  will  only  resist  tension,  while  the 
compressive  chord  strain  due  to  wind  will  sometimes  be  greater 
than  the  tensile  chord  strain  due  to  dead  load,  yet,  as  the 


BRIDGES   OF  CLASS  II.  533 

excess  of  compression  is  not  great,  it  may  be  left  safely  to  the 
outside  longitudinal  I-beams,  which,  it  will  be  remembered,  are 
otherwise  only  half  loaded. 

We  may  compare  the  chord  strains  due  dead  load  and  due 
wind  by  means  of  Fig.  112,  thus  : 

For^F,  N=  ^; 
2n/i 

For  W»  N  =  ^-J-. 
in^h 

These  co-efficients  of  strain  will  be  equal  when 


and  this  might  be  made  a  condition  determining  the  width,  q» 
of  the  bridge,  so  that  no  compressive  strain  would  prevail  in  a 
bottom  chord.  But  we  shall  not  now  change  the  uniform  value 
of  <7,  =  1  8,  assumed  at  first. 

Therefore  the  increase  of  section  of  each  bottom  chord  due 
to  wind,  as  derived  from  (513),  is,  in  square  inches, 


S= 


WJ     \  (n  -  i),a(»-  2),3(«-3),etc.l.    (651) 

0072^,  (  ) 


2000072^ 


The  weight  of  this  wind  augment  to  bottom  chords  is  found 
by  putting  10,000,  instead  of  6,400,  for  Q  in  equations  (515) 
and  (579),  thus  : 


Weight  of  bottom  chords  due  to  wind,  pounds, 

=  o.oo3858o2496(  2  +  -  —  —2  )hl2  (n  even) 

/  ^ 

=  0.00385802496^2  +  |  -  I  -  |JA/'    («  odd) 


(65^) 


534                            MECHANICS  OF  THE   GIRDER. 

=  hi2 

0.0090422 

n  =     8 

=   hi2 

0.0084499 

»   =        15 

0.0088909 

9 

0.0084093 

16 

0.0087963 

10 

0.0083678 

i? 

0.0086957 

ii 

0.0083352 

18 

0.0086272 

12 

0.0083021 

19 

0-0o85555 

13 

0.0082755 

20 

0.0085034 

14 

193.    Assuming  that  a  wind  pressure   per  panel  of    125- 

pounds  (/  being  in  feet)  acts  in  a  direction  normal  to  the  plane 
of  girder  at  each  apex  in  each  top  chord,  we  have  the  moment 
of  each  brace  due  wind  equal  to 


125     X  y  = 


(653) 


if  5  =  cross-section  of  flanges  of  a  brace,  B^  =  inch  strain 
allowed  for  bending-moment  =  £(5,33  3  +  6000)  =  -^f— ,  and 
if  the  length  of  brace  is  to  its  width  at  broadest  end  in  the 
ratio  of  10  to  i,  as  in  article  163,  where  the  flanges  of  each 
brace  meet  at  one  end,  and  diagonal  lattice  work  forms  the 
web.  From  (653), 


Cross-section  of  a  brace  due  wind,  in  square  inches, 

nBl         34/2* 


o 


(654) 


Increase  this  section  by  50  per  cent  for  the  first  and  second 
braces,  to  take  a  part  of  load,  as  explained  in  article  187.  See 
value  of  ^y. 

Weight  of  vertical  braces,  pounds, 


=  2  X  IS  X  34*  X 


=     1.9607844 


(655) 


BRIDGES  OF  CLASS  II. 


535 


=  /// 

2.366729 

n  =  8 

=  hi 

2.091502 

*=   15 

2.291617 

9 

2.076631 

16 

2.235294 

10 

2.064152 

17 

2.192072 

ii 

2-053578 

18 

2.158224 

12 

2.044544 

19 

2.131248 

13 

2.036764 

20 

2.109415 

14 

since  we  now  have  the  special  value 

^     _  4/*  (  (n  —  i)  +  2(n  —  2)  +  $(n  —  3)  -f  .  .  .  (n  —  i)  terms) 
n2  \  -f  (n  —  i)  4-  2(»  —  2)  for  suspenders  j 


n2 


194.  The  Head  Lateral  System.  —  Cross-section  of  head 
diagonals  is  given  by  equations  (587),  (588).     Weight  of  head 
diagonals  found  in  (589)  and  (590). 

Cross-section  of  head  strut  expressed  in  (591)  and  (592). 

Weight  of  head  struts  is  given  by  (593)  and  (594). 

Cross-section  of  iron  to  be  added  to  segments  of  top  chord 
shown  in  (595),  (596). 

Weight  of  added  iron  in  (n  —  6)  panels  of  top  chords  is  to 
be  found  in  equation  (597). 

195.  We  may  now  collect  the  weights  of  all  the  parts  of  the 
bridge,  and,  after  augmenting  by  one-tenth  of  itself  the  weight 
of   the  girders,  the  vertical  braces,   and  the  head   struts,   as 
explained  in  article  165,  we  may  equate  the  weight  so  found  to 
2OOO;/  W,  and  so  determine  W  in  terms  of  L,  /,  and  //,  for  differ- 
ent values  of  ;/,  and  from  — —  =  o  we  find  //  rendering  W  a 

all 

minimum. 


536 


MECHANICS  OF   THE   GIRDER. 


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540  MECHANICS  OF   THE    GIRDER. 

196.  A  few  simple  relations  may  be  stated  here,  as  resulting 
from  our  investigation  of  this  bridge  having  two  parabolic  bow- 
string girders  with  double  triangular  web. 

ist,  For  a  given  uniform  live  load, 

n  oc  I*  nearly,  (669) 

W  oc  Ih  nearly.  (670) 
2d,  For  different  live  loads, 

hW  oc  nL  nearly.  (671) 

197.  EXAMPLE.  —  Specifications  for  the  bridge  of  200  feet 
span,  as  tabulated  in  article  194.     We  have  found 

/  =  200         feet,  n  =  13,                      L  =  15.38460  tons; 

h  =     39.726  feet,  W  =     8.16590  tons; 

q  =     16         feet,  %(W  -{-  L)  —  11.77525  tons; 
g,  =     1 8         feet. 

Maximum  horizontal  component  of  chord  strain,  each  of  2  girders, 

M        n.77525   X  13  X  200 
=  H  =  -J  = 8  x  39.726        -    -  96'333  tons, 

by  equation  (564).     Therefore,  by  (632), 
Strains  in  top  chord  due  loads 

TT 

=  P=  =  119.464  tons,      /.    Sections  =  31.739  square  inches; 

.  cos  a 

=  110.330  tons,  =  29.312  square  inches; 

=  107.231  tons,  =  28.489  square  inches; 

==  102.605  tons,  =  27.260  square  inches; 

=  99. 1 70  tons,  =  26.347  square  inches  ; 

=  97.050  tons,  =  25.783  square  inches; 

=  U               =  96.333  tons,  =  25.593  square  inches; 

for  each  half-span  of  each  girder. 


BRIDGES  OF  CLASS  II.  541 

Additional   cross-section  of   n  —  6  central  panels  of   top 
chords  to  resist  lateral  displacement,  by  (595),  is 

S  =  0.70049  square  inch. 
Corresponding  chord  strain  ==  0.70049  X  3.764  =  2.637  tons. 

/.     Total  top  chord  strains  =  119.464  tons,  ist  and  i3th  panels; 

=  110.330  tons,  2d  and  i2th  panels; 

=  107.231  tons,  3d  and  nth  panels; 

=  105.242  tons,  4th  and  loth  panels ; 

=  101.807  tons,  5th  and    9th  panels; 

=  99.687  tons,  6th  and    8th  panels; 

=  98.970  tons,  7th  panel. 

From  (65 1),  the  varying  sections   of  a  bottom  chord  due 
wind  are 

2500  x  39.7265  x  200 
S  =       20000  x   13;  X  18     ^I2>  2  X  ii,  3  X  10,  etc.) 

=     3.918  square  inches,         .'.     Strains  =  19. 590  tons; 

=     7.183  square  inches,  =  35.915  tons; 

=     9.794  square  inches,  =  48.970  tons; 

=  11.753  square  inches,  =  58.765  tons; 

=  13.059  square  inches,  =  65.295  tons; 

=  13.713  square  inches,  =  68.565  tons; 

Add  19.267  square  inches,  Add  96.333  tons, 

for  total  sections  and  for  total  strains. 
Putting  \L  for  L  in  (647),  we  have 

The  cross-section  of  any  girder  diagonal 

=      3  x  400QO  sec  fl      =   jg  ^ 

64  x  13  x  39.7265 

IST  SYSTEM.  2ND  SYSTEM. 

2d  panel  =  2.250  square  inches  =  3.041  square  inches, 
3d  panel  =  3.791  square  inches  =  3.041  square  inches, 
4th  panel  =  3.791  square  inches  =  4.387  square  inches, 
5th  panel  =  4.795  square  inches  =  4.387  square  inches, 
6th  panel  =  4-795  square  inches  =  5  ooi  square  inches, 
7th  panel  =  5.001  square  inches  =  5.001  square  inches. 


542  MECHANICS  OF   THE   GIRDER, 

The  actual  strains  on  these  girder  diagonals  given  in  the 
diagram  below,  Fig.  120,  where  compressive  strains  are  marked 
negative,  have  been  derived  from  equations  (641)  and  (642), 
using  the  proper  value  of  r,  and  dividing  by  the  proper  value  of 
cos  0. 

The  floor  and  the  longitudinal  I-beams  will  be  the  same 
here  as  in  article  184  ;  viz.,  — 
Floor  of  2^-inch  oak. 
I-beams,  depth  d  =    9.7740  inches. 
d  —  dl  —    0.6610  inch. 
</!  —    9.  1  130  inches. 
S  —    5.0154  square  inches. 
/  =  75.4160. 

Also,  the  cross-section  of  the  transverse  I-beams  due  load 
will  be,  as  in  article  184,  * 

S  =  23.6108  square  inches. 
But,  for  the  wind  pressure, 

2CQO 


x  39-7265  =  7639.7  pounds; 


7.480  square  inches,  /.  Total  =  31.091  square  inches; 

6.285  square  inches,  =  29.896  square  inches  ; 

5.194  square  inches,  =  28.805  square  inches; 

4.207  square  inches,  =  27.818  square  inches; 

3.324  square  inches,  =  26.935  square  inches; 

2.545  square  inches,  =  26.156  square  inches; 


for  each  half-span. 

Take  depth  of  beam  d  =  12  inches. 
d  —  d^  —     2  inches. 
d^  —  10  inches. 
Use  2  I-beams  at  each  joint. 


BRIDGES  OF  CLASS  II.  543 

Then,  by  (558)  and  (557), 

Flange,  b        =  0.178036"  =  5.5351  inches; 

=  5.3224  inches; 

=  5.1282  inches; 

=  4.9525  inches; 

=  4-7953  inches; 

=  4.6566  inches; 
bi      =  0.163636"  =  5.0874  inches  ; 

=  4.8919  inches; 

=  4.7134  inches; 

=  4.5519  inches; 

=  4.4074  inches; 

=  4.2799  inches; 
Web,  b  —  bl  =  0.014406"  =  0.4477  inch; 

=  0.4305  inch  ; 

=  0.4148  inch; 

=  0.4006  inch  ; 

=  0.3879  inch; 

=  0.3767  inch. 

From  (508),  the  cross-section  of  each  horizontal  diagonal  in 
floor  system  is  found,  thus  : 

sin^  =  0.76017, 
7630.7 
5  =  '     x  I2   I2  x  »   "  x  I0  etc' 


=  4.020  square  inches,  ist  and  i3th  panels; 
=  3.402  square  inches,  2d  and  i2th  panels; 
=  2.835  square  inches,  3d  and  nth  panels; 
=  2.319  square  inches,  4th  and  roth  panels; 
=  1.855  square  inches,  5th  and    9th  panels; 
=  1.443  square  inches,  6th  and    8th  panels; 
=  1.082  square  inches,  7th  panel. 
Cross-section  of  head  diagonals  is,  from  (587), 
0.0625  x  168  x  39.7265 


6  x  13'  X  18  X  cos.fr. 

since  cos  fa  =  0.64972. 


°'54"S  ^uare  inch, 


15    X    2OO 

Section  of  a  brace,  by  (654),  =  S  =  -—  -  --  =  6.7873  square  inches. 

34  A  13 


544  MECHANICS   OF   THE    GIRDER. 

Cross-section  of  each  head  strut  is  given  by  (591), 
168  x  39-7265 

S    =  12    X    169          =    3'291   S(1Uare  CS- 

(654),  = 
Add  50  per  cent,  3-3936  square  inches. 

For  2  end  braces,  10.1809  square  inches. 

/ 
RESULTS. 

Use,  in  each  top  chord,  2  lo-inch  channels,  21  square 
inches;  I  1  5-inch  plate  by  0.76  to  0.36  inch;  I  15  X  18  X  J- 
inch  plate  in  every  3  feet,  riveted  to  the  bottom  flanges.  Make 
bottom  chords  of  flat  bars. 

ist  panel,  4  bars,  6  X  0.966  inch  ; 

2d  panel,  5  bars,  6  X  0.882  inch  ; 

3d  panel,  6  bars,  6  X  0.807  incn  \ 

4th  panel,  5  bars,  6  X  1.034  inches; 

5th  panel,  6  bars,  6  X  0.898  inch; 

6th  panel,  5  bars,  6  X  i.ioo  inches; 

7th  panel,  6  bars,  6  X  0.916  inch; 

and  proportion  eyes  as  already  specified. 

In  girder  diagonals,  use  4  angle  irons  2\  X  ^\  inches,  lat- 
ticed both  ways  by  diagonal  strips  of  wrought-iron  \\  X  \  inch, 
and  placed  so  far  apart  that  the  ratio  of  length  to  radius  of 
gyration  shall  be  100,  as  already  provided.  This  will  require, 

40  X   12 
for  a  strut  of  40  feet  length,  a  diameter  of  about  —  ^-     X  2 

=  9.6  inches,  since  by  this  arrangement  of  the  material  the 
radius  of  gyration  is  nearly  one-half  of  the  diameter. 

In  this  case,  where  two  struts  intersect  in  a  panel,  the 
smaller  one  may  pass  within  the  flanges  of  the  larger  one  at 
the  intersection,  involving  thereby  a  little  riveting  in  place, 
and  perhaps  a  little  irregularity  of  the  lattice  work.  The  pin 


BRIDGES   OF  CLASS  II. 


545, 


bearings  at  the  ends  of   these  diagonals  are  to  be  formed  of 
wrought-iron  plates  affording  a  bearing-surface  equal  to 


P. 

I2OOO 


=  2t  X  d, 


(672) 


s 


where  /  —  —  =  thickness  of  plate,  in  inches ;  Px  =  whole 
2a 

pressure  on  strut,  in  pounds ;  d  =  diameter  of  pin,  in  inches  ;  as 
by  general  specifications. 

It  is  here  assumed,  as  in  previous  examples,  that  all  cross- 
sections  can  be  made  exactly  as  the  calculations  require  ;  hence 
we  need  only  notice  further  the  head  struts  and  side  braces. 

For  head  struts,  use  2  fight  4-inch  channels  latticed,  giving 
the  required  section  3.291.  For  wind  braces,  use  2  /-inch 
channels  latticed  with  a  slope  of  I  to  10;  and,  if  a  clear  road- 
way of  more  than  10  (  —  18  —  4  —  4)  feet  is  required,  these 
braces  must  have  their  broad  end  at  the  top,  or  else  they  must 
have  a  bearing  at  the  bottom  beyond  the  ends  of  the  transverse 
I-beams.  Hence  this  mode  of  bracing  high  girders  on  narrow 
bridges  is  objectionable,  and  we  shall  henceforth  either  use  a 
different  style  of  brace,  or  provide  it  a  head  bearing,  or  increase 
the  space  between  girders. 

PARABOLIC  Bow.  —  Two  GIRDERS. 


115.920    132.245    145.305      155.  JO      16.1  ..63      164.895    164.895      32..9S       32.33         31. .02        29.06        26..4S         23..1.8 

FIG.  1 20. 

Span,  200  feet.  Central  height,  39.726  feet  (best) .  Maxima  strains  in  each  girder.  Cross-sections  in 
square  inches.  Live  load,  i  ton  to  the  running-foot.  Bridge  weight,  106.157  tons  (minimum). 
Section  of  bottom  chord  pins  for  shearing  =  3.3  to  3  inches.  Section  of  bottom  chord  pins  for 
bending  —  5.2  to  4  inches.  Diameter  of  bottom  chord  pins  =  33  to  3  inches.  Diameter  of  top 
chord  pins  =  34  inches.  By  equations  (169),  (46),  and  (52). 


546  MECHANICS   OF   THE   GIRDER. 


The  deflection  is  derived  from  (559),  putting  B,  =  ^*.  — 

=  4.3823  tons,  E  —  12,000  tons,  /i,  =  k  =  39.7265  feet,  a  —  \l 
=  100  feet,  and  measuring  x  from  centre  of  span. 

Deflection  /?,  =  1.529  inches  for  x  —        o  at  centre  ; 

D2  =  1.522  inches  for  x  =  100  -^  13  ; 

/)3  =  1.469  inches  for  x  —  300  -^  13  ; 

D4  =  1.361  inches  for  x  =  500  -5-  13  j 

D5  =  1.192  inches  for  x  =  700  -=-  13  ; 

D6  =  0.836  inch      for  x  =  900  -r-  13  ; 

D1  =  0.598  inch     for  x  =  noo  -4-  13  ; 

Z>8  =  o         inch     for  x  =  100  at  end. 

Length  of  top  chord  =  222  x  5  sec  a  =  219.297  feet. 

Contraction  due  strain,  by  (336), 

3.7646  +  5 
=        I200Q  —  x  2I9-297  X  12  =  1.922  inches. 

1.922 
Mean  excess  per  panel  =  -  =  0.1478  =  -^  inch  nearly. 


SECTION  2. 
The  Post  Truss  with  Parabolic  Top  Chord  (Fig.  36). 

198.  Let  our  previous  notation  be  continued  as  far  as  appli- 
cable ;  viz.,  — 

/  =  span,  in  feet. 

//  =  height  of  girder  at  centre,  in  feet. 

n  =  number  of  panels,  counting  on  the  bottom  chord,  and 
odd. 

L  =  panel  weight  of  uniform  live  load,  in  tons,  given. 
W  =  panel  weight  of  bridge,  in  tons,  to  be  determined. 

Symmetry  here  requires  an  odd  number  of  panels  for  the 
bottom  chord,  and  an  even  number  for  the  top  chord.     As  the 


BRIDGES   OF  CLASS  II. 


547 


live  load  will  here  be  applied  to  the  bottom  chord,  we  shall  take 
n  odd,  and  ranging  from  9  to  21,  inclusive.  Each  upper  apex 
is  in  the  middle  of  a  panel's  length.  There  is  but  a  single  sys- 
tem of  counter  diagonals,  while  there  are  two  systems  of  mains. 
We  shall  here  assume  the  difference  of  level  between  the 
centre  and  end  of  the  top  chord  to  be  one-tenth  of  the  whole 
central  height,  h  ;  and  consequently  the  height  at  the  end  of  a 
top  chord  is  -&§h.  This,  of  course,  is  wholly  arbitrary,  except, 
possibly,  in  some  cases  where  the  head  room  at  the  ends  would 
be  too  little.  The  top  chord  is  to  be  polygonal  (that  is,  straight 
from  joint  to  joint),  and  we  will  take  it  parabolic  in  this  case. 


Puttin     2  X 


-/  for  /,  and  —  for  x>  in 


(472),  we  have  the  height  of  any  upper  apex, 


y  = 


Q.^r(n  —  r  —  i) 
(n  -  i)2 


=  «/fc(say);     (673) 


r  to  be  counted  on  top  chord  from  o  to  -      — ,  inclusive  (that 
is,  to  the  centre). 

VALUES  OF  e  IN  (673). 


n  = 

9 

11 

13 

15 

17 

19 

21 

r  =  o 

0.90000 

0.900 

0.900000 

0.900000 

0.900000 

0.900000 

0.900 

i 

0-94375 

0.936 

0.930556 

0.926531 

0.923439 

0.920988 

0.919 

2 

0.97500 

0.964 

0-955556 

0.948980 

0-94375° 

0.939506 

0.936 

3 

Q-99375 

0.984 

0.975000 

0.967347 

0.960938 

0.955556 

0.951 

4 

1.  00000 

0.996 

0.988889 

0.981633 

0.975000 

0.969136 

0.964 

5 

I.OOO 

0.997222 

0.991837 

0.985938 

0.980247 

0-975 

6 

I.OOOOOO 

0-997959 

0-993750 

0.988889 

0.984 

7 

I.OOOOOO 

0.998437 

0.995062 

0.991 

8 

I.OOOOOO 

0.998765 

0.996 

9 

I.OOOOOO 

0-999 

10 

I.OOO 

548 


MECHANICS  OF   THE    GIRDER. 


Since  the  top  chord  for  every  panel  length  slopes  uniformly, 
we  have  manifestly  the  height  of  girder,  if  measured  in  the 
vertical  through  any  lower  apex,  equal  to  the  mean  of  the  two 
heights  at  the  adjacent  upper  apices  just  found. 

If,  then,  we  put  r  +  i  for  r  in  (673),  and  add  the  resulting 
equation  to  (673),  we  have,  after  dividing  by  2, 

(  o.Ar(n  —  r  —  2)  4-  o.2(n  —  2)  ) 

y  =  ;'|°-9  +  -  (n  _  ,).  - }  =  «A  (674) 

which  is  the  height  through  any  lower  apex  ;  r  taking  the  values 


o,  i,  2,  3, 


n  -  3 


,  counted  on  upper  apices. 
VALUES  OF  et  IN  (674). 


«  = 

9 

11 

13 

15 

17 

19 

21 

r  =  0 

0.921875 

0.918 

0.915278 

0.913265 

0.911718 

0.910494 

0.9095 

i 

0-959375 

0.950 

0.943056 

0-937755 

0-933595 

0.930247 

0.9275 

2 

0-984375 

0.974 

0.965278 

0.958163 

o.952344 

0-947  53  i 

0-9435 

3 

0.996875 

0.990 

0.981944 

0.974490 

0.967969 

0.962346 

0-9575 

4 

0.998 

0.993055 

0.986735 

0.980469 

0.974691 

0-9695 

5 

0.99861  1 

0.994898 

0.989844 

0.984568 

0-9795 

6 

0.998980 

0.996094 

0.991975 

0.9875 

7 

0.999218 

0.99691  3 

0-9935 

8 

0.999382 

0-9975 

9 

0-9995 

Calling  a  the  slope  of  any  segment  of  the  top  chord,  we 
have 


snce  \     = 


-  y,.  = 


[»  —  2(r  +  i)]. 


BRIDGES   OF  CLASS  II. 


549 


sec2«  = 


o.i6/i2n2 


x-,, 
i)]2,       (675) 


=  i  +  -«a, 


where  r  is  to  be  counted  o,  i,  2,  3,  etc.,  and 


VALUES  OF  e2  IN  (675). 


n  = 

9 

11 

13 

15 

17 

19 

21 

r  =  o 

O.I55039 

0.156816 

0.157785 

0.158372 

0.158752 

0.159014 

0.159201 

i 

0.079102 

0.094864 

0.105625 

0.113390 

0.119241 

0.123799 

0.127449 

2 

0.028477 

0.048400 

0.063896 

0.075906 

0,085374 

0.092987 

0.099225 

3 

0.003164 

0.017424 

0.032600 

0.045918 

0.057151 

0.066577 

0.074529 

4 

0.001936 

0.011736 

0.023428 

0.034573 

0.044568 

0.053361 

5 

0.001304 

0.008434 

0.017639 

0.026961 

0.035721 

6 

0.000937 

0.006350 

0.013755 

O.O2  1  609 

7 

0.000706 

0.004952 

O.OII025 

8 

0.000550 

0.003969 

9 

O.OOO44I 

199.  Moments  due  a  Total  Dead  Load  of  Uniform 
Panel  Weight,  W  +  L.  —  Although  the  total  load  is  here 
uniform,  the  separate  or  single  systems  are  in  no  case  uni- 
formly loaded  throughout  the  girder's  length  ;  and  we  may  find, 
by  equations  (40)  and  (43),  the  effect  of  each  single  weight, 
IV  +  Z,  at  all  required  points  in  each  single  system,  or  we 
may  sum  the  values  of  a'  in  these  two  equations  for  the  several 
cases,  as  follows  :  — 


ist,  When 
21,  etc. 


—  i 


is  an  integer;  that  is,  when  n  =  9,  13,  17, 


550 


MECHANICS   OF   THE    GIRDER. 


FIRST  SYSTEM.  —  NINE  PANELS. 


y  o        o  o 

'1234  56789 

FIG.  121. 

Let  r  denote  the  number  of   intervals,  each  =  2c  =  —  , 

n 

between  any  weight  in  the  right  half-span  and  the  left  end  of 
the  girder. 

Then,  when  x  7  a',  equation  (43)  applies,  giving 

M=  W  +  La,l_x 


Beginning  at  the  left  weight,  and  summing  the  values  of  a\ 
we  have 


M  = 


r(r 


(676) 


which  is  the  moment  due  all  the  weights  on  the  length,  2cr, 
measured  from  the  left  end,  at  any  point  not  distant  less  than 
2cr  from  the  left  end  of  girder. 

r?^ 


For  the  moments  due  the  weights  on  the  remaining  part, 


/  —  2cr,  we  sum  equation  (40),  where  now  M  = 
and  x  \  a'. 


(/  —  ae)x, 


BRIDGES  OF  CLASS  II.  551 

Beginning  at  the  point  2c(r  +  i),  we  thus  sum  : 

20'  =  2c\_(r  +  i)  +  (r  +  2)  +  (r  +  3)  +  .  .  .  rj 

=  <:(>,  +  r  +  i)(r,  —  r), 
2#'°  =  rx  —  r  =  number  of  terms  ; 

.%     M  =  ^  +  Z[«(r,  -  r)  -  (r,  +  r  +  i)(rx  -  r)]:e 

fX 

=  Ifl±^(«  _  2)-  _  a)(«  _  ar)*,  (677) 


•L  11  —  2 

since  here  r,  =  --  . 

2 

Adding  (676)  to  (677)  gives 


n  -  2T  -  2)(n  -  ar)*,      (678) 


which  is  the  moment  due  all  weights  in  the  first  system,  at  any 

point  in  the  second  half-span,  when  -        -   is  an  integer ;   and 

4 
for  the  panel  points  in  this  system,  second  half-span,  r  becomes 

«  +  i       n  +  5       n  +  9        . 
,       ,       ,      etc., 

444 
and 

2?'/ 

x  =  —  =  2cr. 
n 

Therefore,  putting  this  value  of  x  in  (678), 

M  = (2r  -h  i)(«  —  2r),  (679) 

which  is  the  moment  due  all  the  first  system  weights  at  loaded 
points  in  the  second  half-span. 


552 


MECHANICS  OF   THE   GIRDER. 


If,  in  (678),  x  =  2(r  +  j)-,  it  becomes 

n 


M 


on 


—  9],   (680) 


which  is  the  moment  due  all  weights  at  the  unloaded  panel 
points  in  second  half-span,  first  system. 

Similarly  we  proceed  with  the  second  system  when 


is  an  integer. 


SECOND  SYSTEM.  —  NINE  PANELS. 


(JO  O  O 

1  234567  8  9 

FIG.  122. 


Beginning  at  the  left  weight,  and  summing  the  values  of  a' 
in  (43),  there  results 


n  - 


4- 


.,.,) 


(681) 


which  is  the  moment  due  all  the  weights  on  the  length,  2rc, 
measured  from  the  left  end  of  the  girder,  at  any  point  not  dis- 


BRIDGES  OF  CLASS  II.  553 

tant  less  than  --  from  the  left  end  ;  r  being  not  less  than 

/72   _    j\ 

,  and  increasing  by  unity  for  the  loaded  points  in  second 


half-span. 

For  the  remainder,  /  —  2cr,  we  use  (40),  and  find  equation 

(677),  which,  since  now  rI  —  -  ,  becomes 

M  =  W  +  L(n  -  2T  -  i)2x,  (682) 

4/2 

where  x  cannot  be  greater  than  2c(r  -f-  i),  and  r  not  less  than 
;/  —  i 

4 

Adding  (68  1)  to  (682),  the  result  is,  if,  as  usual,  we  call  the 
sum  M  instead  of  2My 


M=W- 


«-  2T-  i)^,      (683) 


which  is  the  moment  due  all  weights  in  the  second  system,  at 

any  point  in  the  second  half-span,  when  -        -  is  an  integer  ; 

4 
the  limits  of  x  being  2rc  and  2(r  +  i)c>  and  tne  limits  of  r, 

;/  —  i        -,  n  —  i 
-  and  -  . 
4  2 

If,  in  (683),  we  put  x  =  —  ,  we  have,  for  the  loaded  points 

n 

in  second  half-span,  second  system, 


(  W  4- 
M  =  -          ~  [(«  -  *r)(2r  -  i)  +  i]  ;     (684) 


554 


MECHANICS  OF   THE   GIRDER. 


and,  if  x  =  2(r  +  |)->  (683)  becomes 


which  is  the  moment  at  all  upper  or  unloaded  apices  in  second 
half-span,  second  system,  the  sign  of  the  last  moment  to  be 
changed  from  —  to  +. 

2d,  When  -    —  is  an  integer;  that  is,  n  =  n,  15,  19,  etc. 
4 

FIRST  SYSTEM.  —  ELEVEN  PANELS. 


56  789          10        11 


FIG.  123. 

Proceeding,  as  above,  to  sum  a'  in  equations  (40)  and  (43), 
in  the  present  case  we  write 


snce 


where  ^r  7  a'. 


(636) 


BRIDGES   OF  CLASS  IL  555 


Again,  when  x  \  a', 


M  =  W  +  L^(la'°  -  a'}x 


=   W       L(l'1  ~  T)(n  ~  r'  ~  r  ~  X)*'      (687) 


since 

2tf'°  =  rt  —  r  =  number  of  terms, 
and 

20'  =  2t[(r  +  i)  +  (r  +  2)  +  (r  +  3)  +  .  .  .  rj 

=     (r,  -  r)(r,  +  r  +  i). 


Add  (687)  to  (686),  and  put  r,  =  —  ~-~  ; 


-  ar-  2)*,      (688) 


which  is  the  moment  due  all  weights  in  the  first  system,  at  any 

point  in  the  second  half-span,  when  —  -  —  is  an  integer  ;   r 

4 

n  —  I   11  -f-  3  ;/  -}-  7 
being  -  ,  --  ,  -  ,  etc.,  and  x  lying  between  2rc  and 

2(r  +  iy. 

If  x  =  2rc,  equation  (688)  becomes 


(639) 


which  is  the  moment  due  all  the  first  system  weights  at  loaded 
points  in  the  second  half-span. 


556 


MECHANICS  OF   THE   GIRDER. 


If,  in  (688),  x  —  2(r  +  f  )*,  we  have 


(690) 


which  is  the  moment  due  all  weights  at  the  unloaded  apices  in 
the  second  half-span,  first  system. 

Also,  for  the  second  system,  when  —     -  is  an  integer,  we 

4 
write  :  — 

SECOND  SYSTEM.  —  ELEVEN  PANELS. 


U  U 

1234 


From  (43),  we  have 
x  7  a', 


U         U 
567 

FIG.  124. 


n  —  i 


i)], 


11 


-  *)•  (691) 

n 

And,  from  (40), 

#  \  a',     (r,  —  r)  terms ; 
S0'  =  2^r[(r  +  i)  +  (r  +  2)  +  (r  +  3)  +  •  •  •  rj, 


ifr,= 


BRIDGES   OF  CLASS  II.  557 

2tf'°  =  [-(r,  +  r  +  i)(r.  -  r)  +  «(r,  -  r)> 

=  }*•(#  —  2r  —  i) 
—  I 


2 


_  2r  _ 


The  sum  of  (691)  and  (692)  is 

M=  W  +  L\\n  +  i  +  *r(r  +  i)](/  -  x) 

4/1 

+  (n  -  2r  -  i)2^j,     (693) 
which  is  the  moment  due  all  weights  in  the  second  system,  at 

any  point  in  the  second  half-span,  when  -     —  is  an  integer  ; 

4 
the  limits  of  x  being  2rc  and  2(r  +   i)c,  and  the  limits  of  r 

;/  —  3  n  —  i 

being  —  —  and  —  — 

Substituting  2rc  for  x  in  (693),  we  get 


M  =   -  ~~[*  +  i  +  »-(2»  -  4^  -  2)],    (694) 

which  is  the  moment  due  all  weights  at  the  loaded  apices, 

second  half-span,  second  system,  —     -  being  an  integer. 

4 
And,  if  x  =  2(r  +  fX  in  (693),  we  have  finally 


M  =  -  CKS-  -  7)  +  *r(«  -  ar  -  4)],      (695) 


which  is  the  moment  due  all  weights  at  all  upper  or  unloaded 
apices  in  second  half-span,  second  system  ;  the  sign  of  the  last 
moment  to  be  changed  from  —  to  +>  as  in  equation  (685). 

Of  course,  the  total  moments  for  each  and  both  systems 
will  be  equal  at  corresponding  points  in  the  two  half-spans. 


558  MECHANICS  OF   THE   GIRDER. 

200.  Weights  of  Top  and  Bottom  Chords  due  a  Total 
Dead  Load  of  Uniform  Panel  Weight,  W  +  L.  —  Dividing 
(679)  by  (674)  gives 

«3 

(696) 


which  is  the  horizontal  component  of  strain  in  top  chord  over 
loaded  points  in  first  system  if 


(2T 
£,    = 


4* 

and  -  —  -  is  an  integer. 

4 
Also,  dividing  (684)  by  (674),  calling 

(n  —  2r)  (2r  —  i)  4-  i 

6d    =      -  9 

4n 
we  get 


which  is  horizontal  component  of  strain  in  top  chord  over  loaded 
points  in  second  system,  for  this  case.  Then,  adding  the  strains 
due  to  each  panel  length  of  top  chord  from  the  two  systems,  we 
have  total  horizontal  component  of  strain  due  n(W  -f-  L)  in 
each  panel  of  top  chord, 


s  =  x 


if  «5  =  the  sum  of  the  proper  values  of  -  and  -• 

£j  £t 

Similarly,  in  case  —     -  is  an  integer,  making 
4 

_  (n  -  2r)(2r  -  i) 

*'--  -^T 


BRIDGES   OF  CLASS  II. 


559 


in  equation  (689),  and 


(n  —  2r)(2r  +  i)  -f  i 


in  (694). 

Computing  in  this  manner,  we  have,  for  each  half-span, 


VALUES  OF  ss  IN  (698). 


n  = 

9 

11 

13 

15 

17 

19 

21 

Panel  Pt. 

i 

0.704320 

0.679863 

o-74i537 

0.718544 

0.762139 

0.741653 

0.775200 

2 

0.914764 

1.036126 

1.047237 

.122328 

1.121732 

1.174723 

1.169531 

3 

1.120256 

1.204315 

1.341205 

•37989S 

1.468452 

1.487771 

1.55^306 

4 

1.120256 

1.370786 

1.480524 

.629998 

1.690715 

1.791639 

1.828413 

5 

1.370786 

1.621512 

.749982 

1.909621 

1.988183 

2.IOOOOI 

6 

1.621512 

.871858 

2.014554 

2.182537 

2.275878 

7 

.871858 

2.122266 

2.276187 

2.451159 

8 

2.122266 

2.372470 

2.535517 

9 

2.372470 

2.622746 

10 

2.622746 

2e5 

7.719192 

11.323752 

15.707054 

20.688926 

26.423490 

32.775266 

39.864994 

Strain  on  top  chords    =  H  sec  a 


Section  of  top  chords  = 


H  sec  a 

Q 


Weight  of  top  chords,  in  pounds,  due  (  W  -f  Z)  n 


_  5    x  I2/ 

"  18    *   nO 


(^  +  Z)/*x^o^sec^ 
h 

W  +  L  v     10 

, /N 


(699) 


(700) 


since,  by  (675),  sec2  a  =  i  +  j-2s2. 


56o 


MECHANICS   OF   THE    GIRDER. 


Using  the  proper  values  of  e2  and  «5,  we  find 


VALUES  OF  s2ss  IN  (700). 


n  = 

9 

11 

13 

15 

17 

19 

21 

Panel  Pt^ 

i 

0.109197 

0.106613 

0.117003 

0.113797 

0.120991 

o."7933 

0.123413 

2 

0.072376 

0.098291 

0.110614 

0.127261 

0.133756 

0.145430 

0.149056 

3 

0.031901 

0.058289 

0.085697 

0.104742 

0.125368 

0.138343 

0.153928 

4 

0.003545 

0.023632 

0.048265 

0.074846 

0.096626 

0.119282 

0.136270 

5 

0.002654 

0.019030 

0.040999 

0.066021 

0.088609 

0.112058 

6 

0.002114 

0.015787 

0-035535 

0.058843 

0.081297 

7 

0.001754 

0.013476 

0.031309 

0.052967 

8 

0.001498 

0.011748 

0.027954 

9 

0.001305 

0.010410 

10 

0.001157 

2e2f6 

0.434038 

0.578958 

0.765446 

0.958372 

1.186542 

1.425604 

1.697020 

Hence,    if    we    take,    as    in    article    186,    equation    (634), 
Q  =  3.7647  tons,  we  have  finally 


Weight  of  top  chords  due  n(W  -f  Z),  in  pounds, 


W  4-  L 


10 


3  X  3-7647^ 


v7oi) 


o.759412^2  4"  0.042 yoi/*2 
o.9o6448/2  4-  c 


-h  0.0521347^ 
i.22i223/2  4-  0.056571/z2 
I.376226/2  +  o.o6i799/j2 
I.527358/2  4-  0.0664347^ 
i. 68081 1/2  4-  0.07155 i/i2 


n  =  9 
TI 


21 


BRIDGES   OF  CLASS  II.  561 

Again,  dividing  (680)  by  (673)  gives 
H  = 

II  £ 

for  the  bottom  chord  strain  under  an  upper  apex  in  the  first 

n  —  i                                         4r(n  —  2r  —  4)  +  5//  —  9 
system, an  integer,  and  £6  = g • 

Also,  dividing  (685)  by  (673),  and  putting 

g    —  T(11  _  r  _  A  +  **  -  * 
we  have 

H  =  h '    x  I7  (7°3) 

as  the  bottom  chord  strain  under  an  upper  apex  in  the  second 

system  ;  Jl-^  -  being  an  integer. 

4 

Then,  adding  the  two  strains  thus  found  for  each  panel 
length  of  bottom  chord,  we  find,  for  this  case, 


H  =  V"    ^  ">'    X  eg,  (704) 


which  is  the  total  strain  on  bottom  chord  due  n(W  +  L). 

Proceed  in  like  manner  in  case  —  —  is  an  integer,  making 

4 

r/n  \    ,    n  —  •* 

«6  =  -(-  -  r  -  i)  +  — — -* 
n\2  J  8n 

in  (690),  and 


r 

7  =    ("  -  »•  -  4) 


in  (695). 


562 


MECHANICS   OF   THE    GIRDER, 


Computing    thus,    we    find    for    each    half-span,    including 
middle  panel,  — 

VALUES  OF  ss  IN  (704). 


«  = 

9 

11 

13 

15 

17 

19 

21 

Panel. 

i 

0.246913 

0.252525 

0.256410 

0.259259 

0.261438 

0.263158 

0.264550 

2 

0.417791 

0.489510 

0.458860 

0.506825 

0.481072 

0.516987 

0.494988 

3 

0.807154 

0.812868 

0.894161 

0.887470 

0.941122 

0.932223 

0.973214 

4 

0.957264 

1.117273 

1.155222 

1.249844 

1.264131 

1.330849 

I-336554 

5 

i.  linn 

1.238360 

1.408483 

1.471186 

1.578340 

1.613270 

1.689301 

6 

1.363636 

1-509075 

1.687720 

1.770084 

1.888460 

1.940049 

7 

1.615385 

1.774622 

1.960186 

2.058469  |   2.186634 

8 

1.866667 

2.036256 

2.227605 

2-339043 

9 

2.117647 

2.295612 

2.491804 

10 

2.368422 

2.553070 

ii 

2.619047 

2e8 

5.969355 

9.184709 

12.979807 

17.540519 

22.702905 

28.621688 

33.966985 

All  except  the  middle  panel  taken  twice  for  ^£8. 
Taking  T  —  5  tons,  the  allowed  inch  strain  in  tension,  as 
in  (634),  we  find,  from  (704), 


Cross-section  of  bottom  chords,  S  =  (w 


x  £g.       (7o5) 


Weight  of  bottom  chords  due  (  W  +  Z)  n,  in  pounds, 


W+  L 


x       x 

18 


0.442174/2 

O.556649/2 
O.66563I/2 

°-7795  79/2 
0.8903 1  o/2 
I.OO4269/2 
1.07831 7/2 


n  = 


(706) 


ii 

13 
15 
17 
J9 

21 


BRIDGES   OF  CLASS  II.  563 

201.  To  find  the  Greatest  Strains  in  the  Girder  Diag- 
onals, and  their  Weights.  —  Equation  (148)  gives  the  strain 
on  the  counter  diagonals  in  this  case  in  terms  of  the  simulta- 
neous moments  in  the  vertical  planes,  A  A  „  £>£>„  etc.,  Fig.  38. 
These  moments  let  us  compute  for  the  uniform  panel  load,  L 
tons,  advancing  over  the  panel  points  O,  B,  D,  etc.,  of  the 
horizontal  bottom  chord.  The  difference  of  the  horizontal 
strains  due  to  these  moments  at  consecutive  panel  points  will 
be  greatest  when  the  foremost  panel  weight  of  load  is  at  the 
foot  of  a  Y  diagonal  or  counter.  We  use,  therefore,  the  ordi- 
nary formulae  (64)  and  (68),  giving  simultaneous  moments  at  O 
and  B,  B  and  D,  etc.  ;  then,  for  the  moment  at  A  A  „  CClt  etc., 
we  take  one-half  the  sum  of  (64)  and  (68,  thus  : 


=  Z/£9  (say),  (707) 

which  is  the  value  of  Ml  in  (148),  and  would  also  be  obtained 
by  putting  r2  —  o,  x  =  (r  +  $)c  =  (r  +  -|)-,  and  L  for  W,  m 

(60)'  for  the  half-intervals  OA,  BC,  etc.,  Fig.  38. 

Mr+1  in  equation  (68)  takes  the  place  of  M2  in  (148). 

/*,  in  (148)  =  AA19  Fig.  38,    .     =  y  in  (673); 
and 

h2  in  (148)  =  BB{,  Fig.  38,         =  y  in  (674). 

=  i)V+i,  (673)- 


With  these  values  of  Mlt  M2,  hlt  //2,  a»  blt  we  compute  Y  cos 
of  (148),  which,  with  sign  changed,  becomes 

10.  (708) 


564 


MECHANICS  OF   THE   GIRDER. 


VALUES  OF  eIO  IN  (708). 


n  = 

9 

11 

13 

15 

17 

19 

21 

Panel. 

2 

0.024108 

0.016761 

0.012321 

0.009436 

0.007455 

0.006038 

0.004992 

3 

0.063502 

0.044923 

0.033488 

0.025928 

0.020671 

0.016869 

0.014025 

4 

0.115315 

0.082009 

0.061579 

0.048030 

0-038533 

0.031615 

0.026415 

5 

0.180545 

0.127461 

0.095784 

0.074941 

0.060364 

0.049719 

0.041690 

6 

0.182225 

0.136113 

0.106409 

0.086241 

0.070841 

0-059543 

7 

0.183280 

0.142579 

0.114847 

o  094826 

0-079795 

8 

0.183999 

0.147601 

0.121698 

0.102391 

9 

0.184512 

0.151622 

0.127379 

10 

0.184900 

0.154894 

ii 

0.185203 

2e10 

0.766940 

0.906758 

1.045130 

1.182644 

1.320448 

1.456256 

1.592654 

Strain  on  counter 


=  y  =  — -  X  £I0sec<£.    (709) 
fi 


Cross-section  of  counter  =  S  = 


Lh 


Th 


(710) 


Weight  of  (n  —  i)  counters,  pounds, 

o  -  -  -^  — 7^ci< 

1 8  2fi  5/i 

nh 
=  -[  -2«IO  + 


4^2^   A 
""9  / 

0.085  2  1  6/2  +    3-°33912^2 
0.082433/2+    4-386878/z2 


O.O78843/2  +  7.7942407^ 
o.o77673/24-  9-855875Aa 
O.O76645/2  4-  12.140274^ 
0.075841/2  4-  14.6659897^ 


n  =  9 

n 

J3 
15 

17 
19 
21 


BRIDGES   OF  CLASS  II.  565 

since  we  take  T  =  5  tons  per  square  inch  in  tension,  and 


Manifestly  e2  is  to  be  taken  from  (673),  always  beginning 
with  r  =  2. 

Main  Diagonals.  —  To  find  the  strains,  sections,  and  weights 
of  the  main  diagonals  of  the  Post  truss  with  parabolic  top  chord, 
we  proceed  as  follows  :  — 

When  -        -  is  an  integer,  use  equation  (676)  for  the  live 

4 

load,  nL,  making  W  —  o ;  and  for  moment  in  second  half-span, 
first-system  apices, 

At  foremost  end  of  live  load,  put  x  =  ,  giving  M0] 

n 

At  point  \\  panels  ahead  of  foremost  end,  put 

2(7*  -h  J)/ 
x  = — -,  giving  MI', 

At  point  2    panels  ahead  of  foremost  end,  put 

2(r  +  i)/ 
x  = — ,  giving  Ml ; 


these  three  moments  being  simultaneous.     Then 


M0=          « 
«2  (4 


.(713) 


In  a  similar  manner,  for  the  second  half-span,  second-system 


566  MECHANICS  OF   THE   GIRDER. 

apices,  we  find,  from  (68 1),  simultaneous  moments  due  live  load, 


r+I)-  !_![(„  -2r-f)|.      (714) 
,  =  —  jr(r  +  i)  -  ^-^U*  -  2r  -  2) 


Dividing  each  of  these  moments,  (713),  (714),  by  the  height, 
yt  of  truss  at  the  section  where  the  moment  is  taken,  we  find 
the  horizontal  strains^at  the  panel  points  in  second  half-span, 

M 

At  loaded  points,      H0  =  --  from  (713),  (714),  (674); 

Jo 

At  unloaded  points,  H%  =  —^  from  (713),  (714),  (673); 

M, 

At  unloaded  points,  H^  —  —  -  from  (713),  (714),  (674). 

The  difference  of  the  two  simultaneous  horizontal  strains  at 
vertical  sections  through  the  ends  of  a  diagonal  at  and  next 
ahead  of  foremost  end  of  live  uniform  load  is  the  horizontal 
component  of  maximum  strain  on  that  diagonal  due  live  load, 
and  is  tension  on  the  diagonal  whose  foot  is  at  the  foremost 
end,  but  compression  on  the  next. 

LI 
=  N0  -  HZ  =  ^  x  «„  (tension),  (715) 

=  H^  -  HI  =  -^  X  £I2  (compression);     (716) 

£X1  and  el2  being  functions  of  n  and  r  in  (673),  (674),  (713),  (714). 

For  the  moments  due  the  dead  load,  ;/  W,  at  the  same  points 

where  the  simultaneous  moments  due  live  load  have  been  found, 


BRIDGES  OF  CLASS  II. 


567 


—  I 


being  an  integer,  we  use  equations  (679)  and  (680),  and 


(679)  with  r  4-  i  for  r,  L  =  o,  thus  :  — 
First  system, 

MQ  =  —  (2T  +  i)(n  -  2r) 
Wl 


W7 

M,  =  --(2T  +  3)(«  -  2r  -  2) 
4* 

Second  system,  use  (684)  and  (685), 

U/7 

M0  =  --[(«  -  2r)(2r  -  i)  +  i) 


(717) 


=  [(«    -    2T   -    2)(2T 

4/2 


(713) 


Dividing  each  moment  by  the  proper  value  of  y  from  (673) 
and  (674),  we  obtain  horizontal  strains  at  all  required  apices  in 
each  system,  from  the  differences  of  which  consecutive  horizon- 
tal strains  comes  the  horizontal  component  of  maximum  diag- 
onal strain  due  dead  load,  thus  : 

=  H0  —  HI    =  —  X  eI3  (tension),  (719) 

Wl 

=  HI  —  H10  =  X  «I4  (compression);    (720) 

4«/? 

>J3  and  aI4  being  functions  of  n  and  r  in  (673),  (674),  (717),  (718). 


INDEX. 


PAGE 

Advantage  of  steel  over  wrought-iron 246 

Arm  of  resultant  couple 14 

Arm  of  the  couple  defined 10 

Authorities  cited 153 

Bollman  Truss  classified 124 

Bowstring  Girder  classified 101 

Braced  arch 93 

Brunei  Girder  classified 89 

Brunei  Girder  of  double  system 485 

Brunei  Girder  of  double  system,  strain  sheet 521 

Brunei  Girder  of  single  system 423 

Brunei  Girder  of  single  system,  strain  sheet 483 

Camber  calculated  for  certain  cases 281-288 

Camber,  Change  of  length  for 278 

Camber  defined 277 

Camber,  Effective  depth  not  altered  by 286 

Camber,  Length  of  diagonals  changed  by 287 

Cast-iron  pillars 296 

Chord  strains 66 

f  All  members  but  one  inclined 89 

Class  I.  <  Method  of  finding  linear  dimensions 94 

'  One  or  both  chords  parabolic 98 

p.  „  I  Bottom  chord  horizontal.  Other  members  inclined 101 

'  I  Formulae  for 106 

ri  Til  i  Top  chord  horizontal.  Other  members  inclined 109 

'  f  Formulae  for no 

p.  j,r  j  Both  chords  horizontal.  Web  members  inclined no 

'  Formulae  for 113 

(  Both  chords  inclined,  j  Verticals  in  compression  J II4 

Class  V.  <  (  Diagonals  in  tension 

'  Formulae  for 116 

(Both  chords  inclined  j  Verticals  in  tension  j JJ? 

Class  VI.  \  '  Diagonals  in  compression  ) 

v  Formulae  for 118 

569 


570  INDEX. 


PAGE 

(Bottom  chord  horizontal,     j  Verticals  in  compression     j 

Class  VII.  s  i  Diagonals  in  tension 

'  Formulae  for 122 

Cl      VIII   f  Top  chord  horizontal.     Struts  vertical 124 

( Formulae  for 126 

/  T-,   ,,     ,      j    ,      •          .         l  Verticals  in  compression     } 

I  Both  chords  horizontal.  127 

Class     IX.  <  (  Diagonals  in  tension 

'Formulae  for 129 

(Bottom  chord  horizontal,     j  Verticals  in  tension  J 

Class       X.  <  (  Diagonals  in  compression  ) 


*  Formulae  for 131 

ri           XT  I  ^°P  cnord  horizontal.     Struts  inclined.     Ties  vertical 132 

"  '  (  Formulae  for 132 

Class   XII  I  ^ot^  cnorc^s  horizontal.    Struts  inclined.    Ties  vertical 133 

(  Formulae  for 133 

Classification  of  girders 88 

Classification  of  girders,  General  formulae  for 86 

Col.  Merrill's  example  for  Pratt  Truss 399 

Combined  live  and  dead  loads,  Point  of  greatest  moment  due 49 

Component  pressures  or  forces  defined 2 

Composition  of  forces  defined,  with  example 6 

Compound  web  systems 101 

Concentrated  loads,  Formulae  for 19 

Constant  first  difference 37 

Contrary  flexure,  Points  of,  defined 192 

Correction  for  normal  difference  of  moments  due  end  moments 60 

Arm  of,  defined 10 

Arm  of  the  resultant 14 

defined,  with  example 10 

Direction  of  the  resultant 14 

Couples,  Resultant  of  many  co-axal 13 

Crescent  Girder 89 

Deflection  at  centre  of  uniform  beam 172 

Deflection  at  the  free  end  of  parabolic  semi-bowstring 234 

Deflection  of  a  beam,  Influence  of  fixed  ends  on 189 

Deflection  of  a  beam  with  one  fixed  end 192-213 

Deflection  of  girder  of  variable  cross-section 224-277 

Deflection  of  semi-beam 158 

Deflection  of  semi-beam  at  any  interval  due  all  weights 166 

Deflection  of  semi-beam  at  its  free  extremity 162 

Deflection  of  semi-girder  of  uniform  height  and  strength 224 

Deflection  of  uniform  beam  due  any  number  of  equal  weights  at  equal  intervals    .     .     .  183 

Deflection  of  uniform  beam  due  a  concentrated  load 170 

Deflection  of  uniform  beam  due  its  own  weight 169 

Differences,  Method  of  first 36 


Couple, 


INDEX.  571 


PAGE 

Differences  of  simultaneous  moments 36 

Dimensions  of  beam  found 249 

Direction  of  resultant  couple,  with  example 14 

Dome  Principal 116 

Double-bow  or  Brunei  Girder «    .    .    . 89 

Economical  proportions  for  girders,  with  examples 351 

Effect  of  pier  moment 62 

Elastic  curve,  Equation  of 157 

End  moments  accounted  for 60 

Equilibrium  defined 2 

Experimenters  cited 154 

External  forces.     Girder  supported  at  both  ends 23-26 

External  forces  for  semi-girder,  Moments  of 19 

/''defined 312 

Fink  Truss  classified 124 

Floor  beams 314 

Force  defined I 

Forces  acting  at  same  point  in  equilibrium 7 

Forces  in  one  plane 4 

Forces  of  translation  defined 9 

Forces,  Polygon  of,  with  example 6 

Forces,  Resultant  of,  equal  to  zero 6 

G  defined 317 

General  formula?  for  classification  of  girders 86 

Generalized  re-statement  of  locomotive  example 58 

Girders  classified 86 

Girder  supported  at  both  ends.     Moments  of  external  forces  found 23 

Gordon's  formula 299 

Greatest  difference  of  simultaneous  moments 39 

H  denned 87 

h  defined 87 

Hodgkinson's  formulae  for  pillars 296 

Horizontal  girder  of  one  span 59 

Howe  Truss  classified 130,133 

Inertia,  Moment  of 154 

/  defined 314 

Joists,  Proportions  and  weights  of 312 

K  defined 317 

Kansas  City  Bridge  system 105 


5/2  INDEX. 


PAGE 

L  defined 87 

/  defined 24 

Lateral  supporting  system 316 

Linville  Truss  classified 127 

Maximum  deflection  of  a  beam  with  partial  uniform  load 178 

Maximum  moment  at  any  point  due  any  uniformly  continuous  load 48 

Maximum  moment  at  foremost  weight 41 

Maximum  moment  due  dead  and  live  loads,  Point  of 44 

Method  of  first  differences 36 

Modulus  of  rupture 136 

Moment  at  fixed  end  of  semi-beam 19 

Moment  at  foremost  end  a  maximum  for  combined  dead  and  live  loads 48 

Moment  due  both  dead  and  live  load  at  any  point  ahead  of  the  latter 46 

Moment  of  a  force  defined 9 

Moment  of  inertia  defined 154 

Moment  of  resistance  estimated 134 

Moments,  Differences  of  simultaneous 36 

Moments  due  uniform  discontinuous  load 27 

Moments  of  external  forces  found  for  semi-girder 18 

Moments  reversed 31 

Multiple  web  system 101 

P  defined 87 

Parabolic  Bowstring  classified 101 

Parabolic  Bowstring.     Strain  sheet .....545 

Parallelogram  of  forces 2 

Pier  moment,  Effect  of 62 

Pillars 290 

Pillars,  Formula  for.     Case  1 291 

Pillars,  Formula  for.     Case  II 292 

Pillars,  Formula  for.     Case  III 294 

Point  of  greatest  moment  due  both  to  dead  and  live  loads 44 

Point  of  greatest  moment  due  dead  and  live  loads  lying  beyond  the  live  load    ....  45 

Point  of  greatest  moment  due  full  continuous  uniform  load 48 

Point  of  greatest  moment  due  uniform  discontinuous  load 40 

Points  of  contrary  flexure 192 

Points  of  contrary  flexure  for  concentrated  load 196 

Polygon  of  forces,  with  example 6 

Position  of  foremost  end  of  live  load  when  moment  at  that  end  is  maximum   ....  47 

Post  Truss.     Calculations  for  bridge  weight 546-567 

Post  Tiuss  classified 112 

Pratt  Truss.     Best  number  of  panels  and  best  height  found 353 

Pratt  Truss  classified 127 

Pratt  Truss,  Maxima  strains  in 324 

Pratt  Truss  of  single  intersection,  varying  live  load ;  Strain  sheet  for 398 


INDEX.  573 


PAGE 

Pratt  Truss  of  single  system,  uniform  live  load ;  Strain  sheet  for 334 

Pratt  Truss.     Strains  found 318 

Pratt  Truss.     Weight  determined 324 

Pressures  defined I 

r  defined     .     .     .  * 27 

Radius  of  gyration  defined 154 

Rankine's  modification  of  Gordon's  formula 300 

Re-actions  at  the  piers 24 

Resistance  estimated.     Beam  of  elliptical  cross-section 145 

Resistance  estimated.     Beam  of  equal  flanges 137 

Resistance  estimated.     Beam  of  hollow  rectangular  section 137 

Resistance  estimated.     Beam  of  rectangular  cross-section 136 

Resistance  estimated.     Beam  of  two  vertical  channels  and  two  horizontal  plates  .     .     .  140 

Resistance  estimated.     Beam  of  two  vertical  I-beams  and  two  horizontal  plates  .     .    .  139 

Resistance  estimated.     Beam  of  two  vertical  plates  and  two  horizontal  channels  .     .     .  138 

Resistance  estimated.     Solid  or  hollow  beam  of  circular  cross-section 144 

Resistance  estimated.     Solid  or  hollow  beam  of  square  cross-section  and  diagonal  vertical .  142 

Resistance  estimated.     T-shaped  beam 140 

Resistance,  Moment  of 134 

Resolution  of  a  force,  with  example 4 

Resolution  of  many  forces 4 

Resultant  of  forces,  equal  to  zero 6 

Resultant  pressures  or  forces  defined 2 

Rupture,  Modulus  of 136 

Schaffhausen  Truss 131 

Semi-beam  defined 17 

Shearing-force  defined 67 

Shearing-strain  defined 67 

Specifications  for  iron  bridges 413 

St.  Louis  Bridge  system 93 

Strains  deducible  from  moments 64 

Strains  determined  from  shearing- forces 75 

Strain  sheet.     Brunei  Girder  of  single  system 483 

Strain  sheet.     Parabolic  Bowstring  Girder 545 

Strain  sheet.     Pratt  Truss  of  single  intersection,  varying  live  load 398 

Strain  sheet.     Pratt  Truss  of  single  system,  uniform  load 334 

Strains  in  rectangular  beams,  Formulae  for 147 

Table         I.  Ultimate  resistance  of  materials  to  shearing,  in  pounds,  per  square  inch     .  68 

Table       II.  Ultimate  resistance  of  materials  to  tension,  compression,  and  cross-breaking,  149 

Table      III.  Formulae  for  moment  of  inertia  and  square  of  radius  of  gyration  ....  155 

Table      IV.  Values  of  /  and  a  in  the  Gordon  and  Rankine  formulae 302 

Table       V.  Wrought-iron  pillars 304 

Table     VI.  Solid  rectangular  pillars  of  wrought-iron 305 


574  INDEX. 


PAGE 

Table    VII.  Rectangular  tubular  pillars  of  wrought-iron.     Thin 306 

Table  VIII.  Hollow  cylindrical  pillars  of  wrought-iron 307 

Table     IX.  Solid  cylindrical  pillars  of  cast-iron.     Ends  flat 308 

Table       X.  Solid  cylindrical  pillars  of  cast-iron.     Ends  rounded 309 

Table     XI.  Solid  steel  pillars.     Fixed  ends 310 

Table   XII.  Solid  square  pillars  of  pine 311 

Table.     Computation  for  greatest  moments  and  differences 38 

Table.     Computation  for  greatest  moments  and  greatest  simultaneous  differences     .     .  40 

Table.     Deflection  of  open-webbed  girders  of  uniform  strength 242 

Table.     Differences  and  maxima  differences  of  moments  (two  locomotives) 56 

Table  giving  moments  at  any  section  of  beam  supported  at  both  ends 26 

Table  giving  moments  of  forces  applied  to  semi-beam.     Length  / 19 

Table  giving  moments  due  live  load  (two  locomotives) 55 

Table  giving  simultaneous  moments  at  all  panel  points  as  foremost  weight  of  load 

passes  each 43 

Table  giving  simultaneous  moments  due  each  panel  weight  at  panel  point 50 

Table  giving  sum  of  moments  by  (91) 50 

Table.     Horizontal  strains  at  the  joints,  in  tons 402 

Table  showing  best  height  for  least  bridge  weight.     Brunei  Girder  of  double  system     .  508 

Table  showing  least  bridge  weight  for  two  Double  Parabolic  Bow  or  Brunei  Girders     .  467 

Table.     Simultaneous  moments  due  advancing  uniform  load 32 

Table.     Solution  for  dimensions  and  strains.     Method  of  moments 77 

Table.     Solution  for  dimensions  and  strains.     Method  of  shearing-strains 79 

Table.     Strains  deduced  from  moments  and  shearing-forces 83 

Table.     Strains  found  from  moments  for  dead  and  live  loads 85 

Table.    Weight  of  two  locomotives  uniformly  distributed 57 

Thickness  of  beam 246 

Timber  pillars 298 

Triangle  of  forces,  with  example  .    .    .    . 3 

Trussed  rib 115 

Twelve  classes  of  girders 88 

Twin  Fishes 122 

Two  locomotives  as  live  load  on  Pratt  Truss 392 

Two  locomotives,  Discussion  of,  as  moving-load 51 

Typical  form  for  W 360 

U  defined 87 

Uniform  discontinuous  load.     Moment  at  any  weight 35 

Uniform  discontinuous  load.     Moment  at  foremost  end 33 

Uniform  discontinuous  load.     Moment  at  the  rth  weight 34 

Uniform  discontinuous  load,  Moments  due 27 

Uniform  discontinuous  load,  Point  of  greatest  moment  due 40 

Uniformly  continuous  load,  Maximum  moment  due  at  any  point 48 

Uniformly  continuous  load,  Point  of  greatest  moment  due 48 

Uniformly  distributed  loads,  Formulae  for 19 


INDEX.  575 


PAGE 

l\  defined 24 

Vz  defined 24 

v  defined 87 

^defined 87 

W)  Typical  form  for 360 

w  defined 24 

Wind  pressure,  Calculations  for 339-35* 

x  defined    . 24 

Y  denned 87 

y  defined 24 

Z  defined 87 

z  defined 87 


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